'  r 
\ 

I 


I: 


t 


I. 


THE  LIBRARY 

OF 

THE  UNIVERSITY 

OF  CALIFORNIA 

LOS  ANGELES 

GIFT  OF 

Jolm  S.Prell 


■  £'^.:^>  ■  * 

*tf  tit^ti-m^- 


1  «v 


^*,' 


•# 


'  J;?-  '^ 


^'is 


^i 


\..:.  '•/ 


WORKS  OF  PROF.  W.  H.  BURR 


PUBLISHED   BY 


JOHN   WILEY   &   SONS. 


A  Course  on  the  Stresses  in  Bridges  and  Roof- 
trusses,  Arched  Ribs,  and  Suspension  Bridges, 
with  an  Appendix  on  Cantilevers. 

8vo,  500  pages,  13  folding  plates,  cloth,  $3.50. 

Elasticity  and  Resistance  of  Materials  of  Engin- 
eering. 

For  the  use  of  Engineers  and  Students.  Contain 
ing  the  latest  engineering  experience  and  tests. 
6th  Edition,  Rewritten  and  Enlarged.  8vo,  iioo 
pages,  cloth,  $7.50. 

Ancient  and  Modem  Engineering  and  the  isth- 
mian Canal. 

8vo,  XV -|- 473  1  ages,  profusely  illustrated,  includ- 
ing many  half-tones.  Cloth,  $3.50,  net.  Postage, 
27  cents  additional. 


A  COURSE 


THE    STRESSES 


BRIDGE  AND  ROOF  TRUSSES,  ARCHED  RIBS 
AND  SUSPENSION  BRIDGES, 

PREPARED  FOR   THE   DEPARTMENT  OF  CIVIL  ENGINEERING  AT  THE 
RENSSELAER  POLYTECHNIC  INSTITUTE. 

JOHri  S.  PRELL 

Gvil  &  Mechanical  Engineer. 

SAN  FRANCISCO,  CAL. 

By  WILLIAM  H.  BURR,  C.E., 

raoPKSSOR  of  civil   engineering  in  Columbia  college   in  the  city   of  new   york  ; 

MRMBER     OF     THE      AMERICAN     SOCIETY     OF    CIVIL     KNGINEERS. 

NINTH  EDITION  REVISED  AND  ENLARGED. 


FOURTH     THOUSAND, 


NEW  YORK 

JOHN  WILEY  &  SONS 

LONDON 

CHAPMAN  &  HALL,  Ltd. 

1904 


Copyright, 

1886, 

By  VVm.  H.   Burk, 


PRESS  OF 

BRAUNWORTH  &  CO. 

eOOKBINDERS  AND  PRINTERS 

BROOKLYN,  N.  Y. 


^ 


Library 


Id 

1^04- 


PREFACE    TO    THE    EIGHTH    EDITION. 


The  eighth  edition  of  this  work  has  been  considerably 
enlarged  by  a  very  material  extension  of  Arts.  35.  36,  37,  and 
^ya,  on  Swing  Bridges,  which  have  been  entirely  rewritten. 
In  this  revision,  I  am  under  much  indebtedness  to  Mr, 
Henry  W.  Hodge,  C.E.,  for  the  computations  involved  in 
the  numerical  work  of  the  two  actual  cases  of  swing  bridge 
spans  in  Arts.  36  and  T,y.  The  general  theory  of  this  class 
of  structures,  as  first  completely  given  in  the  first  edition  of 
this  work,  has  proved  to  be  so  well  adapted  to  the  rather 
complicated  requirements  of  the  case  that  it  is  now  almo:~t 
universally  adopted  in  this  country.  The  essential  improve- 
ments and  simplifications  presented  in  this  new  matter  make 
the  labor  of  computation  but  little  more  than  that  required 
for  non-continuous  spans. 

W.  H.  B. 

New  York  City, 
/ufy,  1893. 


733584 


PREFACE  TO  THIRD  EDITION. 


Since  the  publication  of  the  first  edition  of  this  book 
•engineering  practice  in  iron  and  steel  construction,  especially 
in  the  department  of  bridge  building,  has  made  very  material 
progress.  The  distribution  of  metal  in  pin  structures  has 
been  considerably  modified  so  as  to  produce  concentrations 
in  larger  members  ;  but  chiefly  the  treatment  of  moving 
loads  has  experienced  such  a  radical  transformation  as  to 
bring  it  to  a  thoroughly  rational  basis.  Hence,  portions  of 
the  book  as  originally  written  have  been  cancelled  and  re- 
placed by  entirely  new  matter,  so  amplified  and  extended  as 
to  bring  the  work  in  all  its  details  abreast  of  the  best  practice 
of  the  present  day. 

My  indebtedness  to  the  published  papers  of  Prof.  H.  T. 
Eddy,  of  the  University  of  Cincinnati,  on  the  arched  rib,  will 
be  evident  to  any  reader  even  slightly  acquainted  with  his 
valuable  work  entitled  "  Researches  in  Graphical  Statics." 

Certain  matters  are  of  such  common  occurrence  in  the  fol- 
lowing pages  that  it  may  conduce  to  clearness  to  mention 
them  here. 

The  word  "  ton  "  signifies  a  ton  of  2,000  pounds,  unless  it 
is  otherwise  specifically  stated. 

The  word  "  stress  "  means  the  force  acting  in  any  member 
of  a  structure,  while  "  strain  "  is  the  distortion  which  accom- 
panies the  stress. 

The  sign  +  indicates  a  tensile  stress,  and  the  sign  — ,  a 
compressive  one. 

Unless  otherwise  stated,  the  stress  in  any  member  of  a 
structure  will  be  represented  by  inclosing  with  a  parenthesis 
the  letter  or  letters  which  belong  to  it  in  the  diagrams  or 
plates.  Thus  (A  B),  or  (a),  signifies  "  stress  in  the  member 
A  B,"  or  "  stress  in  the  member  a." 


VI  PREFACE. 

As  a  matter  of  convenience  to  those  who  may  be  familiar 
■with  the  first  and  second  editions,  it  is  well  to  state  that 
pages  19  to  60,  Arts.  72,  74  and  85  are  entirely  new.  Portions 
of  pages  at  several  other  places  in  the  book  have  also  been 
re-written,  but  it  is  unnecessary  to  name  them  here. 

For  convenience  in  swing  bridge  computations,  Appen- 
dix IV.  has  been  inserted.  Formulae  for  moments  and  re- 
actions are  there  collected  in  the  simplest  form  for  applica. 
tion. 

W.  H.  B. 

Phcenixville,  Pa., 
Feb.  ii,th,  1886. 


'r;  I. 


CONTENTS. 


CHAPTER   1. 

General  Consideration  of  the  Laws  Governing  the  Action  of  Stresses  io 

Trusses. 

PAG? 

Art.     I. — The  Truss  Element i 

"       2. — General   Case 3 

"       3. — Web  and  Chord  Stresses  in  General .  4 

"       4. — Overhanging  Truss — Parallel  Chords — Bracing  with  two  Inclina- 
tions    8 

"       5. — Overhanging  Truss — Parallel  Chords — Uniform  Bracing — Vertical 

and  Diagonal  Bracing II 


CHAPTER    n. 

Special  Non-continuous  Trusses  vrith  Parallel  Chords. 

Art.      6. — Distribution  of  Fixed  and   Moving  Loads 18 

6a. — Trusses  with  Uniform  Panel   Loads 20 

"        7. — Position  of  Moving  Load  for  Greatest  Shear  and  Greatest  Bending.  20z 

' '       8. — Fixed  Weight 36 

"       9. — Single  System  of  Bracing  with  two  Inclinations 38 

"     10. — Single   System  of    Vertical  and   Diagonal    Bracing — Verticals  in 

Tension 50 

"     II. — Single  System  of  Vertical  and    Diagonal    Bracing — Verticals   in 

Compression 54 

"     12. — Two    Systems   of   Vertical    and    Diagonal    Bracing — Verticals  in 

Compression 59 

"     13. — Truss  with  Uniform  Diagonal  Bracing — Two  Systems  of  Triangu- 

lations ....  66 

"     14. — Compound  Triangular  Truss 6g 

"     15. — Methods  of  Obtaining  Stresses — Stress  Sheets 74 

"     16. — Ambiguity  caused  by  Counterbraces 74 


Viii  COh^TENTS. 

CHAPTER   III. 
Non>continuous  Trusses  with  Chords  not  ParalleL 

PAGE 

Art.  17. — General  Methods 75 

"     18. — Curved    Upper  Chord — Two  Systems  of  Vertical  and   Diagonal 

Bracing 78 

"     19. — General  Considerations 88 

*'     20. — Position  of  Moving  Load  for  Greatest  Stress  in  any  Web  Member.     89 

*•     21. — Position  of  Moving  Load  for  Greatest  Stresses  in  Chords 92/ 

"     22. — Horizontal  Component  of  Greatest  Stress  in  any  Web  Member — 
Constant  Value  of  the  Same  for  Vertical  and  Diagonal  Bracing 

with  Parabolic  Chord : :  .     93 

"     23. — Bowstring  Truss — Diagonal  Bracing — Example 93 

*'     24. — Bowstring  Truss — Vertical  and  Diagonal  Bracing  with  Counters.  .    102 
*'     25. — Bowstring  Truss — Vertical  and  Diagonal  Bracing  without  Coun- 
ters     105 

"     26. — Deck  Truss  with  Curved  Lower  Chord,  Concave  Downward — Ex- 
ample      loS 

"     27. — Crane  Trusses iii 

"     28. — Preliminary  to  the  Treatment  of  Roof  Trusses — Wind  Pressure — 

Notation 115 

"     29. — First  Example 117 

"     30. — Second  Example ir8 

"     31. — Third  Example    121 

"     32. — Fourth  Example 122 

"     33. — General  Considerations t . .    123 

CHAPTER    IV. 
Swing  Bridges.     Ends  Simply  Supported. 

Art.  34. — General  Considerations  ....    125 

"     35. — General  Formulae  for  the  Case  of  Ends  Simply  Supported — Two 
Points  of  Support  at  Pivot  Pier — One  Point  of  Support  at  Pivot 

Pier 126 

*'     36. — Ends  Simply  Supported — Two   Points  of  Support  at  Centre — 

Partial  Continuity — Example 133 

"    37. — Ends  Simply  Supported — Two  Points  of  Support  at  Centre — 

Complete  Continuity — Example 159 

"     yja. — Tables  and  Diagrams — Turntables  and  Engines 179 

"     38. — Ends  Simply  Supported — One  Support  at  Centre — Example  ....    187 

CHAPTER    V. 
Swing  Bridges.    Ends  Latched  to  Supports. 

Art.  39. — General  Considerations 203 

"    40. — Ends  Latched  Down — One  Point  of  Support  between  Extremities 

of  Truss — Example 203 


CONTENTS.  ix 

CHAPTER  VI. 
Swing  Bridges.     Ends  Lifted. 

PACK 

Art.  41. — General  Considerations 212 

"  42. — Ends  Lifted — One  Point  of  Support  between  Extremities — Ex- 
ample    212 

"     43. — Final  Observations  on  the  preceding  Methods 222 

CHAPTER  VII. 

Continuous  Trusses  other  than  Souring  Bridges. 

Art.  44. — Formulae  for  Ordinary  Cases — Reactions — Methods  of  Procedure .   224 

CHAPTER    VIII. 

Arched  Ribs. 

Equilibrium  Polygons 229 

Bending  Moments 234 

General  Formulce 236 

Arched  Rib  with  Ends  Fixed 237 

Arched  Rib  with  Free  Ends 247 

Thermal  Stresses  in  the  Arched  Rib  with  Ends  Fixed 250 

Thermal  Stresses  in  the  Arched  Ribwith  Ends  Free 256 

Arched  Rib  with  the  Fixed  Ends — /  and  n  Variable 259 

Determination  of  the  Stresses  in  the  Members  of  an  Arched  Rib — • 

Example — Fixed  Ends — Consideration  of  Details 267 

"     54. — Arched  Rib  Free  at  Ends  and  Jointed  at  the  Crown 278 

CHAPTER    IX. 

Suspension   Bridges. 

Art.  55. — Curve  of  Cable  for  Uniform  Load  per  Unit  of  Span — Suspension 
Rods  Vertical — Heights  of  Towers  Equal  or  Unequal — Gen- 
eralization      271) 

"     56. — Parameter   of  Curve — Distance  of   Lowest  Point  of  Cable  from 

either  Extremity  of  Span — Inclination  of  Cable  at  any  Point. . .    281 

"     57- — Resultant  Tension  at  any  Point  of  Cable 282 

"  58- — Length  of  Curve  between  Vertex  and  any  Point  whose  Co-ordi- 
nates are  x  and  y,  or  at  which  the  Inclination  to  a  Horizontal 
Line  is  ? • 283 

"     59- — Deflection  of  Cable  for  Change  in  Length,  the  Span  remaining  the 

same 285 

60. — Suspension  Canti-Levers 287 


Art.  45. 

' 

46. 

' 

47- 

' 

48. 

■' 

49- 

' 

50. 

•< 

51. 

' 

52. 

' 

53- 

X  CONTENTS. 

PACK 

Art.  6i. — Suspension  Bridge  with  Inclined  Suspension  Rods — Inclination  of 
Cable  to  a  Horizontal  Line — Cable  Tension — Direct  Stress  on 

Platform— Length  of  Cable 288 

"     62. — Suspension  Rods;  Lengths  and  Stresses 292 

"     63. — Pressure  on  the  Tower — Stability  of  the  Latter — Anchorage 294 

"     64. — Theory  of   the    Stiffening    Truss — Ends    Anchored — Continuous 

Load — Single  Weight 296 

"     65. — Theory  of  the  Stiffening  Truss — Ends  Free — Continuous  Load — 

Single  Weight 307 

*'     66. — Approximate  Character  of  the  Preceding  Investigations — Deflec- 
tion of  the  Truss 311 

CHAPTER  X. 
Details  of  Construction. 
A.RT.  67. — Classes   of   Bridges — Forms   of   Compression  Members — Chords 

Continuous  or  Non-Continuous  313 

-Cumulative  Stresses 315 


-  Direct  Stress  Combined  with  Bending  in  Chords • 316 

-Riveted  Joints  and  Pressure  on  Rivets 323 

-Riveted  Connections  between  Web  Members  and  Chords 325 

-Floor-Beams  and  Stringers — Plate  Girders 326 

-Eye-Bars  or  Links 343 

-Size  of  Pins 345 

-Camber 350 

-Economic  Depth  of  Trusses  with  Parallel  Chords 353 

-Fixed  and  Moving  Loads     354 

-Safety  Factors  and  Working  Stresses " 358 

-General  Observations 365 


CHAPTER  XI. 
Wind  Stresses  and  Braced  Piers. 

Art.  80. — Wind  Pressure 366 

' '     81. — Sway   Bracing 370 

"     82. — Transverse    Bracing  for    Transferring  Wind    Stresses  from  One 

Chord  to  Another — Concentrated  Reaction 379 

"     83. — Transverse  Bracing  with   Distributed  Reactions 384 

84. — Stresses  in  Braced    Piers 387 

"     85. — Complete  Design  of  a  Railway  Bridge 395 

APPENDIX  I. 

The  Theorem  of  Three  Moments   431 

APPENDIX    IL 
The  Resistance  of  Solid  Metallic  Rollers   443 

APPENDIX  III. 
The  Schwedler  Truss 447 

-APPENDIX   IV. 

Reactions  and  Moments  for  Continuous  Beams  453 

APPENDIX    V 
Csntilevevs 455 


CONTENTS.  X 

APPENDIX  V. 
Cantilevers. 

PAGE 

Art.  I. — Cantilever  Structures — Positions  of  Loading  for  Greatest  Stresses 

in  the  Cantilever  Arm 455 

"    2. — Positions  of  Moving  Load   for  Greatest  Stresses  in  the    Anchor 

Arm,  and  the  Greatest  Stresses  Themselves 464 

"    3. — Positions  of  Moving  Load  for  Greatest  Stresses  in  Anchor  Spans 

and  the  Greatest  Stresses  Themselves 467 

"    4. — Wind  Stresses 469 

'    5. — Economic  Lengths  of  Spans  and  Arms 472 

*'    6. — Wind  Pressure 474 


CHAPTER    I. 

GENERAL    CONSIDERATION   OF    THE    LAWS    GOVERNING    THE 
ACTION   OF   STRESSES   IN   TRUSSES. 

Art.  1.— The  Truss  Element. 

A  TRUSS  may  be  defined  to  be  a  structure  so  composed  of 
individual  pieces  that,  if  all  the  externally  applied  forces 
called  loading  are  parallel  in  direction,  the  other  external 
forces  called  reactions  will  be  parallel  both  to  each  other  and 
the  loading. 

The  simplest  of  all  trusses  is  a  triangle,  and  all  trusses, 
however  complicated,  containing  no  superfluous  members, 
are,  and  may  be  considered,  assemblages  of  triangles  simply. 
That  the  triangle  is  the  truss  element  arises  from  the  fact 
that  it  is  the  only  geometrical  figure  whose  form  may  not  be 
changed  without  .varying  the  lengths  of  its  sides. 

In  the  elementary  truss  of  the  figure,  let  any  force  act  ver- 
tically downwards  at  B,  and  consider  the  two  triangles  ABD 
and  BDC  having  the  common  side  BD.  Since  all  external 
forces  are  parallel,  the  reaction  at  A  is  to  the  reaction  at  C  as 
DC  x?.  to  AD. 

For  if  BD  be  taken  to  represent  the  vertical  force  acting  at 
B,  and  DF  be  drawn  parallel  to  ^^  as  well  as  EF  parallel  to 
AC,  then  will  DF  represent  the  stress  in  AB,  and  BF  that  in 
BC.  But  by  the  construction  ED  :  EB  =  DC :  AD,  but  ED  is 
the  vertical  component  of  the  stress  in  AB  as  well  as  the  re- 
action at  A,  while  BE  is  the  same  component  of  the  stress  in 
BC,  and,  similarly,  the  reaction  at  C.  It  is  to  be  particularly 
noticed  that  EF  is  the  common  horizontal  component  in  each 
of  the  members  AB  and  BC,  and  also  the  resultant  stress  in 
AC 

I  I 


2  THE  ACTION  OF  STRESSES  IN    TRUSSES. 

When,  therefore,  the  truss  is  horizontal,  as  is  supposed  in 
the  figure,  the  vertical  component  in  each  of  the  members 
AB  and  BC  is  equal  to  the  reaction  at  its  foot ;  also  the  hori- 
zontal component  of  stress  in  each  of  these  members  is  equal 
to  the  horizontal  component  in  the  other,  as  well  as  to  the 
resultant  stress  in  the  third  horizontal  member. 

These  simple  principles  constitute  the  foundation  of  all 
stress  analyses  in  trusses. 


Fig.  I. 


Art.  2. — General  Case. 

Again,  consider  any  truss  whatever,  as  that  in  Fig.  i,  in 
which  the  supports  are  not  on  the  same  level,  nor  are  any  two 
of  the  triangles  of  which  it  is  composed  similar.  Suppose  a 
vertical  load  to  act  at  any  apex,  as  A,  the  reactions  will  be 
vertical.  Let  the  truss  be  cut  by  any  plane  which  divides  the 
line  GH  {AH  is  the  trace  of  such  a  plane),  then  the  part  of 
the  truss  which  is  found  on  the  left  of  AH  is  held  in  equilib- 
rium by  a  component  of  the  vertical  force  at  A,  the  vertical 
reaction  at  C,  and  the  induced  stress  in  GH.  Since  there  is 
equilibrium,  the  lines  of  action  of  those  forces  must  intersect 
in  a  point ;  and  since  the  forces  acting  through  C  and  GH 
have  lines  of  action  intersecting  at  /?,  the  line  of  action  of  the 
component  of  the  vertical  force  at  A  must  pass  through  the 
same  point.  Thus  the  line  of  action  DA  for  one  component 
is  established. 

In  precisely  the  same  manner  BE  is  erected  and  produced 
until  it  intersects  GH,  produced,  in  E  and  the  line  of  action 
AE  of  the  other  component  established.  Connect  D  and  E, 
then,  so  far  as  the  reactions  are  concerned,  the  case  will  not 


GENERAL    CASE.  3 

be  changed  if  the  actual  truss  be  supposed  displaced  by  the 
simple  truss  DEA.  Let  AN  represent  the  vertical  load  at  A, 
then  make  N^O,  parallel  to  AE,  and  DA,  produced,  intersect 
at  O.  If  MO  be  c^rawn  parallel  to  DE,  AM  will  evidently 
represent  the  reaction  at  C,  and  MN  the  reaction  at  B.     Pro- 


duce AN  until  it  intersects  GH  in  G,  and  draw  DK  and  EF 
in  a  horizontal  direction,  then,  from  similar  triangles : 


OM     DG 

MA  =  GA^     ^"^ 

OM       EG 

MN  ~  GA  ' 

ividing  one  by  the  other. 

MN      DG 

DK 

MA  ~  EG 

~  EF' 

But  DK  +  EF  is  equal  to  the  span,  hence  the  reactions  are 

inversely  as  the  segments  of  the  span,  or  any  load  or  system 
of  loading,  vertical  in  direction,  to  which  any  truss  whatever 
is  subjected,  is  divided  into  reactions  according  to  the  law  of 
the  lever.  Farther,  whatever  the  internal  stresses  of  the  truss 
may  be,  at  the  ends  the  sum  of  the  vertical  components  must 
be  equal  to  the  reactions. 


THE   ACTION   OF  STRESSES  IN   TRUSSES. 


Art.  3. — Web  and  Chord  Stresses  in  General. 

In  the  preceding  case  no  account  was  taken  of  the  stresses 
to  which  the  individual  members  of  the  truss  were  subjected, 
but  it  will  now  be  necessary  to  consider  them.  For  this  pur- 
pose PI.  I.,  Fig.  I,  will  be  used,  in  which  the  points  of  support 
will  be  taken  in  the  same  level,  the  loading,  vertical ;  and  the 
truss  will  be  considered  as  made  up  of  similar  triangles.  The 
first  and  last  suppositions  in  nowise  affect  the  generality  of 
the  conclusions,  but  the  operations  are  thereby  simplified  and 
given  a  character  approaching  more  nearly  to  that  of  the 
ordinary  operations  of  the  engineer. 

In  PI.  I.,  Fig.  I,  is  the  representation  of  a  truss  placed  upon 
two  supports  A  and  L  in  the  same  horizontal  line.  CH  and 
AL  are  parallel,  and  the  oblique  members  included  between 
them  may  have  any  inclinations  whatever,  only  they  are  made 
symmetrical  in  reference  to  a  vertical  centre  line  through  O. 
Let  any  weight  W  act  at  any  point,  as  N,  and  let  t  and  /' 
represent  the  tangents  of  the  greater  and  less  inclinations, 
respectively,  of  the  oblique  members  to  a  vertical.  Erect  ver- 
ticals at  A  and  L  which  will  intersect  the  prolongations  of 
CH  in  B  and  K\  then,  as  has  been  shown,  BR  and  UK  drawn 
through  A"  will  be  the  lines  of  action  of  the  components  of  W 
which  act  on  the  two  parts  of  the  truss.  The  force  parallelo- 
gram WRNU  can  then  be  drawn,  in  which  R  T  and  UV  are  to 
be  drawn  parallel  to  AL.  NT=  VW  is  the  reaction  at  A, 
and  vVFthat  at  L. 

Resolve  NR  in  the  direction  of  /TV  and  NM'  by  drawing 
M'R  parallel  to  FN,  then  will  M'R  represent  the  stress  in 
FN.  The  stress  in  FN  will  induce  the  stresses  Fb  and  Fb"y 
in  FG  and  FO,  and  in  the  same  manner  the  stresses  shown  in 
the  figure  will  be  induced  at  all  the  points  on  the  left  of  FN. 

It  is  to  be  noticed  that  all  the  inclined  stresses  at  the  points 
A,  C,  Q,  D,  P,  E,  O,  F,  as  well  as  M'R,  have  the  same  vertical 
component,  NT;  also,  that  all  the  horizontal  stresses  at  C,  Z>, 
F,  and  F  are  equal  to  each  other  and  act  in  the  same  direc- 
tion, each  one  having  the  value  NT  x  {t  +  t').     Let  n  be  the 


WEB  AND    CHORD   STRESSES  IN  GENERAL.  5 

number  of  the  points  C,  D,  E,  and  F,  then  the  total  horizon- 
tal stress  acting  along  CH  from  left  to  right  will  be 

NT  X  (/  +  /')  X  n. 

At  N  there  is  the  horizontal  force  NM'  acting  from  left  to 
right.  Let  d  equal  the  depth  of  the  truss,  or  AB,  and  /  and  /' 
the  segments  ^iVand  NL  respectively  of  the  span  AL  or  s\ 
then  from  the  figure  it  is  seen  that 

NM'  =  NT(^-t\ 

At  the  points  O,  P,  Q,  A,  there  are  horizontal  stresses  act- 
ing from  right  to  left.  From  the  diagram  it  is  seen  that  the 
stress  at  6>  is  (2  /  X  NT)  ;  at  P,  {t  +  /')  x  NT\  at  Q,  {t  +  /')  x 
NT;  at  A,  t'  x  NT\  hence  the  total  stress  on  the  left  of  N 
acting  from  right  to  left  is 


Hence  there  is  deduced  the  important  result  that  NM'  is 
just  equal  to  the  total  horizontal  stress  on  the  left  of  N,  and 
possesses  the  same  line  of  action  but  is  opposite  in  direction, 
therefore  the  two  forces  balance  each  other. 

Next  prolong  GN  until  it  cuts  UV  in  F,  then  will  NY 
represent  the  stress  induced  in  GN  by  the  component  UN. 
The  stress  at  iV,  acting  from  right  to  left,  is  therefore  UY,  or 

UY^NVy.  {^-t 

Although  the  diagrams  are  not  drawn,  it  is  plain  that  the 
horizontal  stresses  at  M  and  L  are  NV  x  (/  -)-  /')  and  NV  x  /' 
respectively  ;  also,  that  their  directions  are  from  left  to  right ; 
hence  the  total  horizontal  stresses,  on  the  right  of  N,  which 
act  from  left  to  right,  are 


6  THE  ACTION  OF  STRESSES  IN    TRUSSES. 

Hence  the  stress  UY  is  balanced  by  the  horizontal  stresses  on 
the  right  of  ^V.  All  the  internal  horizontal  stresses  acting 
along  AL  are,  therefore,  balanced. 

According  to  the  two  force  parallelograms  drawn  from  G 
and  H,  it  is  seen  that  all  the  horizontal  stresses  on  the  right 
of  N,  which  act  from  right  to  left  along  CH,  are 

n'  X  NVx{t  +  t')  =  NVx^-^, 

in  which  n'  is  the  number  of  apices  G  and  H 
B\xtNV=NTj-  hence 

/'  / 

a  a 

But  it  has  already  been  shown  that  all  the  stresses  which 
act  along  CH  from  left  to  right  are 

NT  y^{t  ^t')y.n  =  NT  ^. 

Hence  all  the  horizontal  internal  stresses  of  the  truss  are 
perfectly  balanced  among  themselves. 

This  important  characteristic  belongs  only  to  the  "  truss  " 
proper,  and  distinguishes  it  from  all  other  bridge  superstruc- 
tures. 

If  an  irregular  truss,  like  that  in  Fig.  i  above,  were  treated, 
precisely  the  same  result  would  be  reached,  but  the  resultant 
horizontal  stress  would  be  expressed  by  2Pt  or  2'Pt,m 
which  Pis  a  variable  portion  of  W;  and  ^ would  be  a  "mean  " 
depth,  such  that  2 1 d  =  I,  and  2' td=  /'. 

The  portion  FG  is  subjected  to  all  the  stress  induced  at  the 
points  C,  D,  E,  and  F;  EF  to  all  that  induced  at  C,  D,  and 
E',  etc.  It  is  important  to  notice  this  accumulation  of  stresses 
from  panel  to  panel  in  the  horizontal  lines  CH  3ind  AL,  for 


IVEB  AND   CHORD   STRESSES  IN  DETAIL.  y 

it  shows  that  the  stresses  in  those  portions  are  not  uniform 
from  end  to  end.  A  stress  induced  at  one  point  may  be  felt 
at  any  distance  from  that  point. 

The  upper  and  lower  portions  of  the  truss,  CDEFGH  and 
AQPONML,  are  called  the  top  and  bottom  "  chords,"  and 
all  members  included  between  the  chords,  whether  inclined 
or  vertical,  are  called  "  braces  "  or  web  members.  The  various 
portions  into  which  the  chords  are  divided,  usually  equal  to 
each  other,  are  called  panels. 

From  the  figure  it  is  seen  that  the  vertical  components  of 
the  stresses  in  the  braces  or  web  members  on  one  side  of  N 
are  equal  to  each  other ;  also,  that  the  chord  stresses  have  no 
vertical  component,  being  horizontal.  Farther,  the  vertical 
component  in  any  brace  or  web  member  is  equal  to  the  reac- 
tion found  on  the  same  side  of  the  load  as  itself.  In  a  truss 
provided  with  horizontal  chords,  therefore,  the  ofifice  of  the 
web  members  is  solely  to  transfer,  so  to  speak,  the  load  from 
its  point  of  application  to  the  abutments  or  piers  of  the 
bridge ;  their  duty,  therefore,  is  precisely  the  same  as  that  of 
the  web  in  a  flanged  girder,  hence  their  name  "  web  mem- 
bers." In  other  words,  the  braces  or  web  members  take  up 
the  shearing  stress  at  any  section. 

Let  sec  i  and  sec  i '  be  the  secants  of  the  angles  of  inclina- 
tion of  the  web  members,  corresponding  to  the  tangents  / 
and  t\  and  let  5"  and  S'  be  the  shearing  stresses  in  the  two 
segments  of  the  span  ;  then  the  web  stresses  in  the  left-hand 
segment  will  be  5  sec  i  and  5  sec  i',  and  those  for  the  right- 
hand  segment  5 '  sec  i  and  5  '  sec  i ' . 

The  general  principles  brought  out  in  the  preceding  results, 
therefore,  are  these :  With  horizontal  chords  the  iveb  stresses 
are  products  of  the  shears  by  the  secants  of  the  inclinations,  and 
the  chord  stresses  are  functions  of  the  tangents  of  the  inclina- 
tions of  the  braces  or  web  members  from  vertical  lines. 

From  an  inspection  of  the  force  parallelogram  at  D,  for 
instance,  it  is  seen  that  the  increment  of  chord  stress  at  any 
panel  point  is  equal  to  the  algebraic  sum  of  the  horizontal  com- 
ponents of  the  stresses  in  the  web  members  intersecting  at  that 
point.      The  sum   is  numerical  when,   as  in  the  figure,   the 


8 


THE  ACTION  OF  STRESSES  IN    TRUSSES. 


bfc  ces  slope  on  different  sides  of  the  vertical  line  passing 
through  the  panel  point,  but  the  numerical  difference  is  to  be 
taken  when  both  braces  are  found  on  the  same  side. 

Two  web  members  intersecting  at  any  panel  point  have 
stresses  of  opposite  kindo  induced  by  the  same  shearing 
stress. 

The  stress  in  CH  is  of  course  compressive,  while  that  in 
AL  is  tensile. 

The  preceding  general  results  have  been  deduced  on  the 
supposition  of  the  application  of  but  one  weight,  but  they 
are  equally  true  for  any  system  of  loading.  For  the  effect  of 
any  system  of  loading  is  simply  the  summation  of  the  effects 
of  the  individual  loads  of  which  it  is  composed,  hence  only 
those  principles  which  are  true  for  the  individuals  can  be  true 
for  the  system,  and  those  at  least  must  hold,  for  the  action  of 
each  load  is  independent  of  all  the  others. 


Art.  4. — Overhanging  Truss. — Parallel  Chords. — Bracing  with   Two  In- 
clinations. 

Probably  the  simplest  case  of  a  truss  subjected  to  the  action 
of  external  loading  occurring  in  the  practice  of  the  engineer 
is  a  simple  truss  fixed  at  one  end,  and  is  the  case  with  one 
arm  of  a  swing-bridge  when  open  and  subjected  to  its  own 
weight  as  load. 

Now,  in  all  cases  of  actual  trusses,  the  load  will  be  sup- 
posed divided  in  its  application  to  the  truss  between  the 
upper  and  lower  chords.  It  will  not,  however,  be  equally 
divided,  because  the  floor  system  of  the  bridge  will  rest 
wholly  on  one  chord  or  the  other. 

h  c  d  e  f    T 


Fig.  I. 


In  the  figure,  let  the  truss  be  fixed  at  AB,  and  let  W  and 


OVERHANGING    TRUSS. 


W  be  the  panel  loads  on  the  upper  and  lower  chords  respec- 
tively, except  at  the  extremity  P  of  the  upper  chord,  where 
h  W  will  rest.  Let  a  represent  the  angle  QPR,  and  a'  the 
angle  SQT\  the  line  PR  is  vertical,  and  y^/^  horizontal. 

A  simple  and  direct  application  of  the  principles  and  for- 
mulae of  Art.  3  gives  the  following  results  : 


Stress 

in  0 

"   M 

"    K 

"    G 

"   E 

"   D 

"    N 

''    L 

"   H 

"   F 

-    V 

"    C 

ec  a. 

{ihW 

+ 

(2i  VV 

+ 

2W') 

{3vy 

+ 

1>^V') 

« 

(4yV 

+  aW) 

isvv 

+ 

SlV) 

(  W 

+ 

IV')- 

sec  a! 

{qw 

+ 

2  IV) 

{2lW 

+ 

3in 

(3U^ 

+ 

4  IF') 

{Ah^V-^ 

SW) 

{ShW 

+ 

6W') 

It  will  be  observed  that  in  every  instance  the  stress  in  any 
brace  is  the  "  shearing  stress "  in  the  section  to  which  the 
brace  belongs,  multiplied  by  the  secant  of  the  inclination  to 
a  vertical  line  of  the  brace  in  question.  For  instance,  if  the 
brace  L  be  divided  by  a  vertical  plane,  the  weights  on  the 
right  of  it,  or  the  shearing  stress,  are  {^\IV  -^  2 IF'),  and  this 
multiplied  by  the  secant  of  a  is  the  stress  desired. 

The  chord  stresses  are  also  determined  by  a  direct  and  sim- 
ple application  of  the  principles  and  formulae  of  the  preceding 
article. 


Stress  in  /  is    \  IV  tan  o: 

"       "    e  "  i^lVtan  a-\-(lV'  +  ^W)  (tan  a  +  fan  cr') 
"       "  ^  "  2,\  IV tan  a +  {2  IV+  3  IV)  (tan  a  +  tan  a') 
"       "    c  "  zWtan  r*'  +  (4.i  IF-I-6IF')  (tan  a^tan  a) 
"       "   b  "  4^  Wtan  n'  +  (8  W^  10  W ')  {tan  a  +  tan  a) 
"       "  ^  "  5 ^  W  tan  a  +  {\2\  IV+  1 5  IV)  (tan  a  +  tan  a') , 


(0- 


In  the  lower  chord  the  stresses  are  as  follows  : 


10  THE  ACTION  OF  STRESSES  IN    TRUSSES. 

Stressing  is     W  tana' ■^\W  {tan  a-'rtana'^  ~1 

"  "  h''  2W'  tana'  +  {2lV+W'){tana+tana')         | 

u  u  j^u  ^Wfana'  +  {4lW+T,lV'){tana  +  tana') 

-     "  "  /  ''  4W' tana'  +  {SW+6W'){tana  +  tan  a') 

u  "f^*'  ^W tana +{i2lW+ioW') {tana  +  tana)  \ 

u  a  ^u  eW  tana'  +  {iSlV+isW'){tana  +  tana')) 


K2). 


In  determining  these  quantities,  it  is  to  be  remembered  that 
the  stresses  cumulate  from  the  free  end  to  the  fixed  ;  i.e.,  the 
stress  developed  at  any  panel  point  is  felt  throughout  those 
portions  of  the  chords  included  between  that  point  and  the 
fixed  end  of  the  truss. 

General  formulae  for  the  Eqs.  (i)  and  (2)  may  easily  be 
found.  Let  n  be  the  number  of  the  panel,  from  the  free 
end,  in  the  chord  AP {/is  number  i  ;  r,  2  ;  ^,  3  ;  etc.),  then 
the  formula  expressing  the  results  in  Eq.  (i)  is  the  following: 

Stress  in  any  panel  =  {n  —  ^)  W  tan  a  -\-  \  - — —  W  + 

-^^ '  W   \  \tan  a  +  tan  a\        ....     (3). 

This  expression  gives  the  stress  in  any  panel  of  AP. 

The  formula  which  expresses  the  results  shown  in  Eq.  (2) 
is  the  following : 

Stress  in  any  panel  =  nW  tan  a'  +  {  —  W  + 

-^ W    ^  \tana  +  tana'\    .     .     .     (4). 

In  which  n  denotes  the  number  of  the  panel  in  the  chord  BQ 
starting  from  the  free  end  ;  i.e.,  g  is  number  i,  h  number  2, 
etc. 

The  weight  at  P  has  been  taken  at  half  that  applied  at 
other  panel  points  in  the  same  chord.  In  the  case  of  a  swing- 
bridge,  however,  it  is  greater  than  that,  since  some  of  the 
details  of  the  locking  apparatus,  etc.,  are  hung  from  that 
point.     Yet  the  Equations  (i)  to  (4)  may  still  be  used,  only  a 


OVERHANGING    TRUSS.  \\ 

simple  term  is  to  be  added  to  each  of  those  equations.  Let 
/  be  the  panel  length,  d  the  depth  of  the  truss,  and  IF,  the 
actual  weight  hung  from  P.  Also,  let  Wy—\W=w'.  In 
order  to  find  the  additional  stress  produced  in  any  panel  d  of 
the  chord  AP,  let  the  moment  of  w  be  taken  about  the  in- 
tersection of  H  and  K  in  the  lower  chord  ;  this  moment  is 
w\{n —  \)p -{- d  tail  a\.     Consequently  the  additional    stress 

desired  is 

I  p  \ 

s— w^An— \)^-\-  tana  r    .     .     .     .     (5). 

The  stress  s  is  to  be  added  to  each  of  equations  (i)  and  (3) 
if  Wi>  ^  W,  otherwise  it  is  to  be  subtracted. 

In  precisely  the  same  manner,  the  additional  stress  for  the 
lower  chord  BQ  is 

s'  =  wn  ^ (6). 

d 

The  stress  s  is  to  be  added  to  equations  (2)  and  (4)  if  fFj  > 
\  W,  otherwise  it  is  to  be  subtracted. 

If,  as  in  Fig.  i,  AP  is  the  upper  chord,  the  stress  in  QP 
and  all  members  parallel  to  it  will  be  compressive  ;  while  the 
stress  in  QS  and  all  braces  parallel  to  it  will  be  tensile.  Like- 
wise the  stress  in  AP  is  tensile,  and  that  in  BQ  compressive. 

If  the  truss  were  turned  over  so  that  ^^  would  become  the 
top  chord,  the  expressions  for  the  stresses  in  equations  (i)  to 
(6)  would  remain  exactly  as  they  are,  only  the  signs  of  the 
stresses  would  change.  The  condition  of  stress  would  be  ex- 
actly represented  in  the  preceding  paragraph  by  simply  chang- 
ing "compressive  "  to  "  tensile,"  and  "  tensile  "  to  "  compres- 
sive." 

Art.  5. — Overhanging    Truss — Parallel    Chords — Uniform   Bracing — Ver- 
tical and  Diagonal  Bracing. 

The  two  most  frequent  cases  of  Fig.  i,  Art.  4,  are,  first, 
that  in  which  a  —  a',  and,  second,  that  in  which  a'  —  o.  The 
first  of  these  cases  is  represented  in  Fig.  i,  and  the  second  in 
Fig.  2. 


12 


THE  ACTION  OF  STRESSES  IN    TRUSSES. 


The  web  stresses  for  this  case  will  be  precisely  the  same  in 
general  form  as  those  given  in  Art.  4,  but  sec  a  will  be  written 
for  sec  a. 

Very  simple  general  formulae  can  be  written  for  these  web 
stresses.  Let  n'  denote  the  number  of  any  brace  starting 
from  O,  which  is  called  i  ;  then  observing  the  general  values 
in  Art.  4,  the  stress  in  any  brace  71'  parallel  to  O  will  be 

+  d  =  \  ^  W  +  ^^   ~ — -  IV  i  sec  a  +  w  sec  a.    .    .    (i). 


Fig.   I. 


The  expression  (+  d),  of  course,  denotes  tensile  stress  in  any 
brace  parallel  to  O. 

In  precisely  the  same  manner,  the  compressive  stress  in  any 
brace  parallel  to  A^  {n  possessing  the  same  signification  as 
before  ;  i.  e.,  n  for  A^  is  2  ;  for  L,  4,  etc.)  is 

,        (  («'  —  i)  „.       n    „.,  )  ,  , 

—  b  —  \ IV  +  —  W   >  sec  a  +  w  sec  a    .    .    (2). 


In  determining  the  chord  stresses,  it  is  to  be  remembered 
that  the  weights  IV  rest  on  the  lower  chord  AP.  Making 
tan  a  —  tan  a'  in  Eq.  (3)  of  Art.  4,  the  stress  in  any  lower- 
chord  panel  is 

C—  {n-h)lVtana-\-  \  {n—\yW+n{n—\)W'  \  tana-\-s  .  (3). 


Making  the  same  change  in  Eq.  (4)  of  the  previous  Article, 
the  upper-chord  tensile  stresses  will  be  found  to  be 


T—71W'  tan  (x  +  \  n^  f F+ ;/(;/—  i )  f f '  \  tan  a  +  s 


(4). 


0  VEKHA  NGING    TR  USS. 


13 


Some  of  the  results  given  by  the  formulae  should  always  be 
checked  by  the  method  of  moments. 

Let  it  be  desired  to  determine  the  stress  in  k  by  the  method 
of  moments.  Let  the  origin  of  moments  be  taken  at  the 
intersection  of  G  and  H.  The  moments  which  balance  each 
other  about  that  point  are  that  of  the  stress  in  k  acting  with 
the  lever-arm  d,  the  depth  of  the  truss,  and  those'  of  the 
weights  applied  to  the  truss  on  the  right  of  the  panel  point 
in  question  ;  these  latter  act  against  the  former.  Calling  the 
panel  length/,  and  taking  the  moments  mentioned: 

7^  =  3  W  •  1/  +  2  W  •  \p  ^  \W  •  IP  ^-  w  '  ip. 

:.   T=4VV'^  +  4W^+3^^    •    •    •    (5). 

The  result  of  Eq.  (5)  ought  to  be  the  same  as  that  of  Eq. 
(4).  Two  or  three  panels  in  each  chord  ought  to  be  treated 
in  the  same  manner. 


Fig.  2. 


The  cantilever  truss  represented  in  Fig.  2  shows  the  rase 
in  which  m'  of  Fig.  i,  Art.  4,  is  equal  to  zero.  The  note-fion 
is  precisely  the  same  as  that  used  before. 

The  general  expression  for  the  stress  in  any  inclined  brace 
is  simply  Eq.  (i)  repeated — that  is: 


+  d 


i  2 


I) 


JT'  [  sec  a  +  IV  sec  a 


(6). 


Making  sec  a'  =  i,  there  results  for  the  compressive  stress 
in  any  of  the  verticals  2,  4.  6,  etc. : 


14 


THE  ACTION  OF  STRESSES  IN    TRUSSES. 
-I) 


b  =  i  ^— 2— ^  ^¥+-^'1  +  ^ (7). 


Making  faft  a  =  o,  in  Eq.  (3)  of  Art.  4,  gives  the  com- 
pressive  chord  stress  in  any  panel  of  the  lower  chord  AP. 
Hence,  for  that  chord  : 


C 


i                 ,,-       (n  —  i)^  „.      n  (n  —  i)  „.,  , 
=  j  («  -  i)  ^V  +  ^- — ■—  W  +  — ^^ ^  W  \   tmi  a  + 


wn 


P 


(8). 


In  a  similar  manner,  from  Eq.  (4)  of  the  previous  article, 
for  the  tensile  stress  in  any  panel  of  the  upper  chord,  there 
results  the  equation : 


T  = 


IV  + 


ti  {n  —  i) 


W  \  tan  a  +  wn 


(9). 


If  Wi  should  be  less  than  \W,  the  term  which  expresses 
the  additional  stress,  whether  in  braces  or  chords,  will  be  sub- 
tractive,  as  will  be  indicated  by  the  sign  of  w. 


Fig.  3. 

Again,  applying  the  moment  test  to  any  panel,  as  c,  by 
taking  the  origin  of  moments  at  R,  the  notation  remaining  the 
same  as  before,  there  results : 


.'.C^{4W+slV')^^+4^^ 


(10). 


OVERHANGING    TRUSS. 


15 


This  result  ought  to  agree  with  that  shown  by  Equation  (8), 
and  several  panels  in  each  chord  should  be  tested. 

In  the  great  majority  of  cases  it  is  not  convenient  to  apply 
a  general  formula,  but  the  numerical  values  are  usually  deter- 
mined directly  from  the  diagram,  and  the  stress  in  each  mem- 
ber written  along  it,  as  shown  in  Fig.  3. 

Fig.  3  shows  a  truss  which  frequently  occurs  in  the  practice 
of  the  American  engineer ;  it  is  in  reality  one  arm  of  an  open 
swing-bridge. 

Let  the  panel  length  =  p  =  \2  ft.,  and  the  depth  of  the 
truss  =  d  =  20  ft.  The  tangent  of  CQP  =  12  ^  20  =  0.6,  and 
the  secant  of  CQP  —  1.166.  The  tangent  of  DEP  =  1.2,  and 
the  secant  of  DEP—  1.562.  The  panel  loads  at  E,  F,  H,  etc. 
=  W^'  =  3-00  tons;  at  C,  D,  G,  K,  etc.,  W=  5.00  tons,  at 
P,  IVi  =  4.00  tons.     No  load  is  taken  at  Q. 

In  the  figure  there  are  two  pystems  of  right-angled  triangu- 
lation  ;  P,  E,  D,  H,  K,  etc.,  is  one  system,  and  C,  F,  G,  L,  M, 
etc.,  is  the  other.  This  does  not,  however,  complicate  the 
matter  in  the  least,  for  each  system  of  triangulation  is  regarded 
as  ail  individual  truss  carrymg  its  own  weights  only.  Calcula- 
tions are  therefore  made  for  each  system  of  triangulation  as 
if  they  were  independent  trusses,  and  then  the  two  are 
added. 

The  weight  of  the  portion  EQP  is  supposed  divided  be- 
tween E  and  P,  thus  showing  W^  >  i  W.  The  vertical  braces 
are  evidently  in  compression,  while  the  inclined  ones  are  in 
tension. 

The  figures  in  the  diagram  denote  tons  (2000  lbs.)  of  stress; 
-I-  indicates  tension,  while  —  indicates  compression. 

Stress  in  PE  =  IV^  see  DEP  —  4  x  1.562  =    6.248  tons. 
"       "    CF  =W  sec  DEP  ^t^  ^  1.^62^    7.810    " 
"       "   DH={U\  +  W  +  lV)x  1.562  =  18.744    " 
"       "    GL=  2PV+lV'x  1.562  =  20.306    " 

The  other  brace  or  web  stresses  are  found  in  precisely  the 
same  manner. 


l6  THE   ACTION  OF  STRESSES  IN    TRUSSES. 

Stress  in  CP  —  W^  tan  DEP  =4      x     1.2 

=  4.8  tons 
"       "  DC  =  Wta?i  DEP  +  4.8  =    6.0  +  4.8 

=  10.8  tons. 
"       "  GD=  { W\  +  IV  +  W)  X  1.2  +  10.8  =  14.4  +  10.8 

=  25.2  tons. 
"       ''  KG  =  {W^-  W   +  W)x  1.2  +  25.2        =  40.8  tons. 

Other  lower-chord  stresses  are  found  in  exactly  the  same 
manner. 

By  an  inspection  of  the  diagram  it  is  seen  that  the  general 
expression  for  the  stress  in  FE  is  precisely  the  same  as  that 
for  CP\  the  same  can  be  said  oi  HE  \n  reference  to  DC;  LH 
in  reference  to  DG ;  etc.  The  explanation  of  this  is  simple. 
If  the  truss  be  divided  by  a  plane  normal  to  the  paper  and 
parallel  to  the  inclined  braces,  only  vertical  and  horizontal 
members  will  be  cut.  But  the  truss  is  in  equilibrium,  and 
since  the  loading  is  wholly  vertical,  the  sum  of  the  horizontal 
stresses  must  be  zero ;  or,  the  stress  in  the  lower-chord  panel 
cut  must  be  equal  and  opposite  to  the  stress  in  the  upper- 
chord  panel  cut. 

Let  the  moment  test  be  applied  to  the  stress  in  the  panel 
MK.  The  origin  of  moments  for  the  loads  applied  to  the 
system  PEDH,  etc.,  is  N,  and  the  origin  for  the  other  system 
is  L.     Taking  moments  about  those  points : 

C'd  +  C"d  =  16  X  36  +  4  X  72  +  8  X  24  -h  5  X  48 

.-.  C  —  C  -\-  C"  =  64.8  tons. 

Again,  for  the  lower-chord  panel  adjacent  to  A,  B  is  the 
moment  origin  for  the  whole  load. 

C=  {48  X  42  +  5  X  84  +  4  X  96)  ^  20  =  141  tons. 

Thus  the  numerical  results  are  verified. 

According  to  one  of  the  principles  of  Art.  3,  the  horizontal 
component  of  the  stress  in  any  inclined  web  member  ought 
to  be  equal  to  the  increment  of  chord  stress  at  either  of  its 


0  VERHA  XG/.VG    TR  USS. 


17 


extremities,  and  such  will  be  found  to  be  the  case.  If,  for 
example,  20.3  be  multiplied  by  the  cosine  of  the  angle  GLH, 
the  result  will  be  15.6  =  40.8  —  25.2. 

This  last  is  a  verification  of  the  web  stresses,  and  both 
methods  of  checking  are  perfectly  general  and  may  be  applied 
to  all  trusses,  as  should  be  done  in  actual  cases. 


CHAPTER    II. 

SPECIAL  NON-CONTINUOUS   TRUSSES   WITH    PARALLEL 
CHORDS. 

Art.  6. — iDistribution  of  Fixed  and  Moving  Loads. 

The  trusses  treated  heretofore  have  been  of  rather  an  ele- 
mentary character,  and  general  principles  have  been  con- 
sidered instead  of  special  and  practical  applications.  Before 
taking  up  the  technical  treatment  of  trusses  it  will  be  neces- 
sary to  consider  some  preliminary  matters. 

The  total  load  on  a  bridge-truss  always  consists  of  two 
parts,  the  fixed  load  and  the  moving  load.  The  fixed  load 
consists  of  the  entire  weight  of  the  bridge,  including  tracks, 
flooring,  etc.  The  moving  load,  as  its  name  indicates,  con- 
sists of  that  load  (whether  single  or  continuous)  which  moves 
over  the  bridge. 

The  truss  is,  of  course,  always  subjected  to  the  action  of 
the  fixed  load. 

If  the  truss  is  of  uniform  depth  the  panel-fixed  loads  will 
be  uniform  in  amount  for  one  chord  ;  but  if  the  depth  is  vari- 
able it  may  be  necessary  to  make  a  varying  distribution  of 
the  weight  of  the  trusses  and  lateral  bracing.  The  amount 
and  rate  of  this  variation  can  only  be  determined  by  the  cir- 
cumstances of  each  particular  case. 

The  moving  load  on  a  railway  bridge  may  be  taken  as  con- 
tinuous or  as  a  series  of  single  weights  as  actually  applied  at 
the  wheels  under  the  locomotives  and  cars.  The  assumption 
of  continuity  of  moving  load  was  formerly  always  made,  a 
larger  amount  per  lineal  foot  being  taken  to  represent  the 
extra  locomotive  w^eight.  In  such  a  case,  if  the  moving  load 
extends  from  the  end  of  the  bridge  to  the  centre  of  any 
panel,  or   to  the  end  of  that  panel,  the  panel  point  immedi- 

i8 


RAILWAY  BRIDGES.  19 

ately  in  front  of  the  train  will  not  sustain  a  full  panel  load  ; 
but  if  it  be  assumed  that  this  panel  point  does  sustain  the 
full  load,  then  a  small  error  on  the  side  of  safety  will  be 
committed.  Such  an  assumption  was  formerly  made,  and 
the  consequent  method  of  computation  will  be  given  in  some 
of  the  Arts,  which  follow  in  this  chapter. 

At  the  present  time  (1885),  however,  the  demands  of  the 
best  practice  require  the  moving  load  to  be  taken  at  the 
actual  points  of  application  of  locomotive  and  car  wheels. 
This  method  of  computation  will  be  given  in  several  of  the 
first  cases  taken. 

If  the  span  is  short,  or  less  than  125  feet,  the  moving 
load  should  be  taken  entirely  of  locomotives,  as  two  will 
nearly  cover  the  structure.  The  amount  and  character  of 
the  moving  load,  however,  is  usually  indicated  by  specifica- 
tions. 

The  moving  load  of  a  bridge  may  pass  along  the  upper 
chord  or  the  lower  chord.  In  the  first  case  the  bridge  is 
called  a  "  deck  "  bridge,  and  in  the  latter  case  a  "  through  " 
bridge.  The  methods  employed  in  the  determination  of 
stresses  in  the  various  truss  members  are  exactly  the  same  in 
both  cases. 

"  Pony  "  trusses  are  through  trusses  not  sufficiently  high  or 
-deep  to  need  overhead  cross-bracing. 

Every  truss-bridge  is  composed  of  the  following  parts: 

Upper  and  lower  chords. 

Upper  sway-bracing, 

Web  members, 

Floor  system,  including  beams,  stringers,  ties,  floor-hangers, 
lower  sway-bracing,  and  rails. 

The  sum  of  the  w'eights  of  the  parts  is  the  "  fixed  "  load  of 
the  bridge. 

In  the  case  of  highway  bridges  the  calculations  are  pre- 
cisely the  same  as  for  railway  bridges,  except  that  the  mov- 
ing load  is  assumed  to  be  uniform  per  lineal  foot  of  bridge. 
The  greatest  load  that  can  ordinarily  pass  on  a  bridge  is  a 
dense  crowd  of  people,  the  greatest  w'eight  of  which  can  be 
taken   at  eighty-five  pounds  per  square  foot.     The  late  Mr. 


20 


SPECIAL  NON-CONTINUOUS   TRUSSES. 


Hatfield,  of  New  York  City,  found  by  experiment  that  it  was 
scarcely  possible  to  exceed  seventy  pounds  per  square  foot. 
The  moving  panel  load  of  a  highway  bridge  may  then  be 
found  by  multiplying  the  width  of  the  clear  way,  including 
sidewalks,  by  the  product  of  the  panel  length  with  the  load 
per  square  foot. 

If  the  span  is  not  over  125  feet,  or  about  that  value,  the 
moving  load  for  the  truss  members  may  be  taken  at  eighty- 
five  pounds  per  square  foot,  or  sixty  pounds  for  greater 
lengths.  In  all  cases,  however,  the  floor  beams  and  joists 
should  be  designed  for  a  moving  load  of  100  pounds  per 
square  foot,  in  order  to  provide  for  the  increased  fatigue  of 
those  members  due  to  shocks  and  sudden  application  of 
loads. 

In  some  cases  highway  bridges  are  subjected  to  enormous, 
concentrated  loads  of  a  special  character.  Such  loads  can 
only  be  known  from  local  considerations,  and  the  bridges 
must  be  built  with  a  view  to  sustaining  such  special  weights. 

All  the  methods  or  principles  used,  then,  in  the  following 
cases,  which  will  be  those  of  railway  bridges,  are  equally  ap- 
plicable to  highway  structures,  and  no  further  special  atten- 
tion will  be  given  to  the  latter. 

Art.  6  a.  — Trusses  with  Uniform  Panel  Loads. 

Fig.  I  represents  a  simple  truss  of  the  Pratt  type,  sup- 
ported at  each  end  on  an  abutment,  and  supposed  to  carry 
a  system  of  uniform  moving  panel  loads  in  addition  to  its 


own  weight  or  fixed  load.  The  moving  load  will  be  taken 
as  traversing  the  lower  chord.  Although  the  truss  has  but 
nine  panels,  the  formulae  will  be  written  so  as  to  apply  to 
any  number  of  panels. 


7RUSSES    WITH    UNIFORM  PANEL   LOADS.  20a 

The  inclined  web  members,  except  the  end-posts,  are  in 
tension.  The  end-posts,  vertical  web  members  (except  the 
two  Zi's),  and  upper  chord  are  in  compression,  while  the 
lower  chord  is  in  tension.  The  two  vertical  tension  mem- 
bers, Tx,  are  really  not  members  of  the  truss,  but  they  are 
virtually  hangers,  to  carry  the  loads  at  their  feet  to  the  upper 
chord  panel  points  at  their  tops.  The  moving  load  is  sup- 
posed to  pass  on  the  bridge  from  right  to  left,  and  the  nota- 
tion is  so  written  as  to  apply  to  any  number  of  panels. 

li\  —  uniform  lower  fixed  panel  load. 
w     —  uniform  lower  moving  panel  load. 
W  —  uniform  upper  fixed  panel  load. 
/      =  panel  length. 
/      =  length  of  span. 

a     =  uniform  angle  of  inclination  of  inclined  mem- 
bers to  a  vertical  line. 

It  will  be  convenient  and  more  simple  to  treat  the  fixed 
and  moving  loads  separately. 

Case   I, — Fixed  Loads. 

The  inclined  members  marked  c^,  c^,  etc.,  and  shown  by 
broken    lines,   will    be   considered    omitted,   as   they  are  not 

stressed  by  the  fixed  or  dead  loads.     If  iV  = i   is  the  total 

P 
number  of  upper  chord  panel  points,  or  the  total  number  of 
lower  chord  points  at  which  loads  are  applied,  the  reaction 
R  due  to  the  fixed  loads  only  will  be  : 


K^^SK^J^ ^^^ 


Then,  if    n   be    any    of   the   subscript    numbers   of   c^,  c^, 
Ci,     .     .     .      T-;,  the  shear  in  any  of  the  members  T  will  be  : 


S  =  R  -{N-  n)  {W  +  W,)  =  n{W,  +  W) 


(2). 


20b  SPECIAL  NON-CONTINUOUS   TRUSSES. 

Both  the  second  and  third  members  of  Eq.  (2)  show  that 
the  fixed  load  shear  in  any  main  web  member  Z„  is  equal  to  the 
total  fixed  load  between  the  centre  of  the  span  and  the  member 
T,^.  If  the  number  of  panels  in  the  span  is  odd,  this  princi- 
ple shows  that  the  shear  in  the  inclined  web  members,  or 
diagonals,  of  the  centre  panel  will  be  zero  ;  and  it  also  shows 
that  if  the  number  of  panels  is  even,  the  shear  in  each  of  the 
diagonals  running  upward  from  the  centre  panel  point  of  the 
lower  chord  will  be  |  {W  +  Wy). 

The  stress  in  any  inclined  main  web  member  T^  will  there- 
fore be : 

(7;,)  =  S  sec  a  ^  {W  +  W,)  (n  -  ^\  sec  a  .     (3). 
The  stress  in  any  vertical  post  P^  will  be  : 

(/'„):=(^^'+^0(«-f)     +     ^'         .        .        (4). 

N 
In  these  formulae  n  will  never  be  less  than  — . 

2 

The  stress  in  the  inclined  end-post  EP  will  be : 

(^EP)  =  ^ -^  seca,     ,    .     .     (5). 

The  stress  in  Zi  will  evidently  be : 

(^1)=  ^1 (6). 

Inasmuch  as  the  arrangement  of  the  truss  members  is 
symmetrical  about  a  vertical  line  through  the  centre  of  the 
span,  it  is  necessary  to  find  the  stresses  in  the  members  of 
one-half  the  truss  only. 

When  loads  are  so  symmetrically  disposed  about  the  centre 
of  the  truss  that  the  inclined  web  members  c,,  c^,  etc.,  which 
are  shown  as  broken  lines  in  Fig.  i,  and  are  called  "  counters  " 
or  "  counterbraces,"  are  not  stressed,  they  (the  loads)  are  said 
to  be  "  balanced."  All  uniform  fixed  or  dead  loads  in  sym- 
metrical trusses  are  balanced. 


TRUSSES    WITH    UNIFORM  PANEL   LOADS.  20C 

The  fixed-load  chord  stresses  remain  to  be  determined. 
In  accordance  with  the  principles  already  established,  the 
lower  chord  stress  in  panels  i  will  be  the  shear  or  vertical 
component  in  £/*  multiplied  by  tan  a.     Hence  : 

{l)  =  ^^  {W  +   l]^,)tan  a  .     .      .     .     (7). 

Since  the  stress  in  71  has  no  rectangular  horizontal  com- 
ponent, the  two  stresses  in  the  lower  chord  panels  i  will  be 
identical. 

The  stress  in  lower  chord  panel  2  will  be  equal  to  the  shear 
in  T^f-i  multiplied  by  ta7i  a,  added  to  (i).  Hence,  by  the 
aid  of  Eq.  (3),  in  which  ji  —  N  —  i  : 

(2)  =  ^  ( W  +   W,)  tancY  +  (^  -  i)  ( ^V'  +   W,)  tan  a 
=  -(iV  -  I)  ( ^F'  +   W^  tan  a     .     .     .     (8). 

Similarly,  the  stress  in  lower  chord  panel  3  is  equal  to  the 
shear  in  7v_2  multiplied  by  tan  a,  added  to  (2).  Hence,  by 
aid  of  Eq.  (3),  as  before,  in  which  ;/  is  now  equal  to  iV—  2  : 

(3)  =  (A^-  I)  {W  +  W,)  tana  +  (~-  2)  {W  +  W,)  tan  a 

=  ^(jV- 2}{IV'  +  W,)tana       .     .     .     (9). 
In  a  similar  manner: 

(4)  =^{N-s){W'  +  W,)  tana   .     .     (10). 

The  second  member  of  Eq.  (7)  and  the  third  members  of 
Eqs.  (8),  (9),  and  (10)  show  that  if  n'  indicate  the  number  of 
any  lower  chord  panel,  as  exhibited  in  Fig.  i,  the  general 
expression  for  any  lower  chord  stress  will  be  : 

n'  =z  -  {N -  (n' -  i)\  (IV  +  W,)  tan  a    .     (11). 


20^!" 


SPECIAL   NON-CONTINUOUS    TRUSSES. 


If  the  number  of  panels  in  the  truss  is  even,  the  greatest 
— ^— "^l    but  if  that  number  is  odd,  the 


value  of    ri  will  be 


greatest  value  of 


■n  K     ^ 
will  be  — . 

2 


The  upper  chord  will  evidently  be  subjected  to  compression, 
and  the  stresses  in  that  member  can  be  written  immediately 
from  the  lower  chord  stresses.  If  the  truss  be  divided  through 
any  main  panel,  as  a2,bi,  c4,  etc.,  it  is  at  once  clear  that  the 
only  horizontal  stresses  in  the  section  are  those  in  the  two 
chord  panels  cut,  since  the  posts  that  are  cut  are  vertical, 
and  have  no  horizontal  components  of  stress.  The  upper  and 
lower  chord  stresses  are,  therefore,  equal  in  pairs,  but  of 
opposite  signs.     Hence  : 

(«)  -  -  (2) 

(^)  =  -  (3) 

(^)  =  -  (4) 

etc.  =  —  etc. 


!^ 


(12). 


In  the  case  of  an  even  nurpber  of  panels  in  the  truss.  Fig. 
2,  the  stress  in  the  centre  pair  of  panels  of  the  upper  chord 
will  be  greater  than  the  result  given  by  Eq.  (ii)with  the 
greatest  value  of  n',  by  the  amount  ^  {W'  +  W^)  tan  a.  This  is 
for  the  reason  that  two  inclined  main  web  members  intersect 
at  the  middle  point  of  the  lower  chord,  while  their  upper  ends 
include  the  centre  pair  of  panels  of  the  upper  chord.  The 
horizontal  component  of  the  stress  in  each  of  these  inclined 
main  web  members  \s  h  {W  -{-  W^  tan  (y. 

If  the  end-posts  of  the  truss  are  vertical,  as  shown  in  Fig. 
2,  the  stress  in  the  end  lower  chord  panel,  o,  is  zero  ;  but  the 


Fig.  2. 


stresses  in  the  panels  1,2,  3,  etc.,  are  given  by  Eq.(i  i)  precisely 
as  before.     Then,  also,  in  the  same  general  way  as  before : 


TRUSSES    WITH    UNIFORM  PANEL   LOADS.  2(V 

(^)--(2)     V.      ....      .      (13). 

etc.  =  —  etc.  ) 

The  expressions  for  the  web  stresses  will  remain  unchanged 
•and  as  given  by  Eqs.  (3)  and  (4).  As  the  end-post,  EP,  is 
now  vertical : 

{EP)  -  -  {W'  +  Wy)  +  W    .     .     .     (14). 

The  central  vertical  member,  P^,  is,  strictly  speaking,  not  a 
truss  member  for  the  fixed  load  ;  hence  its  stress  due  to  fixed 
load  is  simply  the  amount  of  that  load  applied  at  its  end. 
No  formula  is  to  be  applied. 

Case  II. — Moving-  Loads. 

.  The  moving  load  will  be  taken  as  passing  on  the  span, 
panel  by  panel,  from  A  towards  B,  Fig.  i.  When  the  first 
panel  load  is  at  the  foot  of  c,  the  second  will  be  at  the  foot 
of  Ti.  Hence  the  stress  in  c,  will  be  the  shear  in  that  mem- 
ber, /.  e.,  w  ~  +  w  -J-,  multiplied  by  sec  oi\  ox\ 

P 
(c,)  =  {i  +  2)w  J  sec  a        .     .     .     (15). 

When  the  head  of  the  train  is  at  the  foot  of  c^  there  will  be 
panel  loads  at  the  foot  of  /\  and  71,  and  the  shear  in  c-i  will  be 

(zv  ~  +  w  ~  +  w  ^j  ;  hence : 

(cs)  =  {l  +  2  +  ^)  wj  sec  a    .     .     .     (16). 

Similarly,  for  any  other  inclined  web  member  (except  the 
^end-post)  with  the  subscript  n  : 

{T„)  =  (i  +  2  +  .     .     .  +  n)w  y  sec  a  .    .     (17). 


20/  SPECIAL  NON-CONTINUOUS    TRUSSES. 

The  general  expression  for  the  sum  of  the  series  in  the 
parenthesis  gives: 

{T,)  =  n{^yvo^jseca      .     .     .     (i8). 

As  the  shear  in  any  incHned  web  member  is  equal  to  the 
compression  in  the  post  whose  upper  end  it  meets  in  the 
upper  chord : 

(/-„)  =  n  (^)  «.  f (.9). 

As  Zi  is  simply  a  hanger : 

{T,)  =  w    ,    ,     .     ,     ,    ,     (20). 

The  stress  in  EP  is : 

N 
{EP)  =  —  w  sec  oc (21). 

The  stress  in  any  lower  chord  panel  whose  number  is  w'can 
be  taken  directly  from  Eq.  (11)  by  placing  w  for  {W'  +  Wi). 
Hence : 

(«')  =  —  \N  —  {n'  —  i)]  zu  tan  a     .     .     (22). 


The  upper  chord  stresses  wiU  be  found  from  the  lower,  as 

indicated  by  Eq.  (12);  and  the  greatest  value  of  ;/'  will  be 

N                                                                                           N  —  I 
-when  the  number  of  panels  in  the  span  is  odd  and 

when  that  number  is  even. 

If  the  end-posts  are  vertical,  as  shown  in  Fig.  2,  the  relation 
shown  by  Eq.  (13)  still  holds,  while  the  web  stresses  remain 
unchanged.     The  vertical  end-post  stress  becomes: 

{EP)='^w (23> 


TJiUSSES    WITH    UNIFORM  PANEL   LOADS. 


20g 


Case  W.a. — Moving  Loads — Deck  Span. 

If  the  moving  load  traverses  the  upper  chords,  as  indicated 
in  Fig.  3,  the  post  {P)  subscripts  shown  in  Figs,  i  and  2  are 
to  be  increased  by  one,  so  that  the  post  and  tie  meeting  in 
the  lower  chord  will    have  the  same  subscript.     Fig.  3  ex- 


FiG.  3. 

hibits  the  notation  required.  The  expressions  for  the  stresses 
in  the  main  ties  and  posts,  T  and  P,  will  be  precisely  as  given 
in  Eqs.  (18)  and  (19) : 


(  7;j  =  «  ( )  w  -J  sec  a 


{Pn)  =  n 


n  4-  I 


w 


(24). 


In  the  cases  where  a  counter  member,  c,  acts  the  part  of  a 
post,  there  is  a  little  ambiguity  in  the  post  stress,  and  it  may 
be  advisable  to  make  it  equal  to  the  shear  in  T^-x  in- 
creased by  w.     In  that  case: 


(i^„)=|(..-  1)^^+1}^. 


No  other  change  is  needed  in  the  formulae  for  the  through 
truss  to  meet  the  "  deck"  condition,  except  that.  1\  will 
have  no  moving  load  stress,  and  if  the  end-post  is  vertical  Eq. 
(23)  will  take  the  form : 


(^-)=(f-:) 


w. 


(25). 


As  there  is  more  or  less  concentration  at  a  panel  point  with 


2Qk  SPECIAL   NON-CONTINUOUS    TRUSSES. 

an  actual  moving  load,  it  is  usually  advisable,  in  the  case  of 
vertical  end-posts,  to  take : 

{EP)  =  (^  +i)w....     {2Sa). 


Case    III. —  Wed    Members    zvith     Two    Inclinations — Fixed 

Loads. 

This  case  is  represented  by  Fig.  4,  in  which  the  end-posts 
and  the  main  web  members  parallel  to  them,  i.  e.,  the  web 
members  in  compression,  have  the  inclination  a' ;  while  the 
main  web  members  in  tension,  such  as  1 -^  to  T^,  have  the 
inclination  a'  to  a  vertical  line.  Inasmuch  as  the  truss 
should  be  symmetrical,  however,  the  members  P^  and  7^  of 
the  centre  panel  must  have  the  same  inclination,  a,  to  a 
vertical  line.  Strictly  speaking,  therefore,  the  web  members 
have  three  inclinations. 

For  the  reasons  explained  in  the  preceding  cases,  the  fixed 
loads  will  be  balanced,  and,  hence,  there  will  be  no  stresses 
corresponding  to  those  loads  for  the  members  Zi  to  7^,  inclu- 
sive, and  Px  to  P^,  inclusive,  or  for  members  corresponding  to 
them  for  a  different  number  of  panels  from  that  shown. 


Inclination  of  EP  to  vertical  —  a. 
Inclination  of  T»  to  vertical  =  a" . 
Inclination  of  Pt  or  Ti  to  vertical  =  a. 


As  before,  W  will  be  the  upper  chord  panel  fixed  load, 
and    W^i  the  lower  chord  panel  fixed  load.     Also,  /  =  panel 

length,  /=  length  of  span,  and  N  = i  =  total  number  of 

loads  sustained  by  the  truss  on  the  lower  chord.     The  number 


TRUSSES    WITH    UNIFORM  PANEL  LOADS.  20i 

of  loads  sustained  in  the  upper  chord  will  be  {N  +  \).     Hence, 
the  reaction  at  either  A  or  B  will  be : 

jV  A^  4-  I 

R,  =  -  U\  +  '—^^^  W  .     .     .     .     (26). 


By  using  the  same  process  employed  in  establishing  Eq.  (2), 
the  shear  in  any  member,  T,  with  the  subscript,  n,  will  be: 

S=n{W'  +  IV,)  -  f^lVr  +  '^^^— -^  W'^ 
=  (^n-'-^y_lV'+lV,)-'^    ....     (27). 


The  second  parenthesis  in  the  second  member  will  b^- 
recognized  as  the  reaction  R . 

Similarly,  the  shear  in  any  member  P  with  the  subscript  n 
will  be  : 

IIV  N  +  I       \ 

S'  =  n{W'  +  W,)  +  W  -  r^  W,  4-  — ^  W'\ 

=  {n-^){^'  +  m+^-^ ^^^)- 

Hence,  the  stress  in  any  member  T  between  the  centre 
panel  and  the  end  of  the  truss,  i.  e.,  in  Fig.  4,  from  T^  to  T^, 
inclusive,  is : 

i^n)=    \{n-^y,W'  +W,)-^lseca"     .     .     .     (29), 

Similarly,  for  any  member  P  between  the  centre  panel  and 

the  end  of  the  truss : 

iPn)=-    ][n-^){lV'  +  W,)  +  ^\seca'.     .     .     (30). 

N 
And  for  the  centre  panel,  for  which  n  =  —  : 
^  2 

w 

{Pd  =  m=--^seca  ....     (31). 


lOJ  SPECIAL   NON-CONTINUOUS    TRUSSES. 

If  the  number  of  panels  in  the  lower  chord  is  even,  so  that 

a  panel  point  is  at  the  centre  of  the  span,  there  will  be  no 

members  with  the  inclination   a,  and    Eq.  (31)  will    not   be 

needed  ;   Eqs.  (29)  and  (30)  will  then  give  all   the  fixed  load 

web  stresses.     Eq.  (30)  will  give  the  end-post  stress  (^P),  by 

N 
giving  ;/  its  proper  value,  — .     The  equation  then  becomes : 

fN  W'\ 

{EP)  =  ('J  {W  +   W,)  +  -^  j  sec  a  .     .      (32). 

The  chord  stresses  due  to  the  fixed  loads  can  best  be  written 
by  the  method  of  moments.  If  the  number  of  any  lower 
chord  panel  be  indicated  by  ;/,  the  origin  of  moments  for  the 
stress  in  that  panel  will  be  the  panel  point  over  it  in  the 
upper  chord,  and  the  distance  of  the  moment  origin  from 
the  nearest  abutment  (the  lever  arm  of  the  reaction  R^  will 
be  (;/'  —  \)  p  +  d  tan  a',  ^^  being  the  depth  of  the  truss.  The 
distance  from  the  centre  of  the  {ri!  —  i)  loads  in  the  lower 
chord  to  the  same  origin  will  be  \  [n'  —  2)  p  +  d  tan  a  ;  and 
the  distance  from  the  centre  of  the  (>/  —  i)  loads  in  the  upper 
chord  to  the  same  origin  will  be  \  n'p.  The  lever  arm  of  («') 
will  be  d.  The  equation  of  moments  about  the  origin 
described  will  then  be : 

{n')d=  R,  I  («'  -  i)p  +  dtan  a  \  -  (;/'  -  i)W, 

|i(V  -  2)p  +  dtana']^  -  {n'  -  i)W'  "^  p .     .     (33). 

By  substituting  the  value  of  R^  from  Eq.  (26)  and  then 
dividing  both  members  by  d,  remembering  that/  =  d {tan  a 
+  tan  a"),  the  following  expression  for  («')  will  result: 


(«')  =  ^  ^iN-n'  +  i){lV'  +  IV,)  +  IV' 


(tan  a'  +  tan  a") 


+    I  («'  -  i)W,  -  ^{W,  +  W)  -  ^  I  tan  a"  .     (34). 


7-A' (/SSES    WITH   UNIFORM  PANEL   LOADS.  20k 

The  moment  origin  for  any  upper  chord  panel,  U^,  is  at  the 
panel  point  underneath  it  in  the  lower  chord,  distant  ?ip  from 
the  abutment.     The    lever  arm  of  the  {n  —  i)  lower  chord 

ft'f) 

panel  loads  will  be  — ,  and  that  of  the  n  upper  chord  panel 

loads  will  have  the  value  ^ ~  +  d  tan  «".    The  equation 

of  moments  will  then  be : 

{U^)d  =  R,np  -  {n  -  i)W,  f  -  nW  [(«  -  i)^ 

+  dtan  a-"  I (35)- 

By  inserting  the  value  of  R,  from  Eq.  (26)  and  remembering 
that  p  =  d  {tan  a'  +  tan  a")  : 

Un  =  -  \{N-  n  +  \)  {W  +  W,)  +  W     {tan  a'  +  tan  a") 

—  nW  tan  a" (36). 

Eqs.  (34)  and  (36)  will  give  all  the  chord  stresses  in  the 
truss  except  that  in  the  centre  lower  chord  panel,  which  will 
be  called  {m).  The  fact  that  the  two  web  members  over  it 
are  equally  inclined  to  a  vertical  makes  a  separate  expression 
necessary  for  that  panel.  By  taking  moments  about  the 
centre  panel  point  of  the  upper  chord  there  will  result; 

(m)  d  =  Ri '-^  —  ~Wi—p -W 

^    ^  2  24^2 


4 

Hence : 


l^^dtana'^ (37). 


(^)  =  [^(^+^)  ( PT'  +  FTO  +  ^^  ^']  {tan  a'  +  tan  a") 


-  ^I'V  tan  a" (38). 

2 


20/  SPECIAL   NON-CONTINUOUS    TRUSSES. 

Eq.  (38)  will  give  the  stress  in  any  centre  lower  chord 
panel,  when  there  is  one.  It  will  evidently  not  be  needed 
when  the  number  of  panels  in  the  lower  chord  is  even. 

In  the  application  of  Eq.  (34)  the  greatest  value  of  n  will 

N  .  N  +  \ 

be  —  for  an  odd  number  of  panels  in  the  span,  and for 

2  ^  ^  2 

an  even  number. 

If  the  end  posts,  EP,  are  vertical,  tan  a  becomes  equal  to 
zero,  but  no  other  changes  are  required  in  any  of  the  preced- 
ing formulae. 


Case  IV. — Case  III.,  with  Moving  Loads. 

As  heretofore,  w  will  represent  the  uniform  panel  moving 
load.  The  latter  will  be  taken  as  passing  from  A  to  ^  in 
Fig.  4.  All  notation  will  remain  as  in  Case  III.  With  the 
progress  of  the  moving  load  from  A  to  B  the  stresses  which 
it  produces  in  all  the  members  marked  7i,  T.^,  etc.,  will  be  ten- 
sile, but  those  in  P^,  P-i,  etc.,  will  be  compressive.  Between 
A  and  the  centre  of  the  bridge,  therefore,  the  moving  load 
will  produce  stresses  in  the  web  members  of  a  kind  opposite 
to  those  caused  by  the  fixed  load.  If,  then,  the  moving  load 
stress  of  one  kind  in  any  member  is  greater  than  the  fixed 
load  stress  of  the  other  kind  in  the  same  member,  that  mem- 
ber must  be  so  designed  as  to  sustain  both  tension  and  com- 
pression, i.e.,  it  must  be  countcrbraccd.  The  resultant  moving 
load  stress  in  any  web  member  of  the  half  of  the  truss  first 
traversed  by  that  load  will  be,  then,  the  excess  of  the  great- 
est latter  stress  over  that  produced  by  the  fixed  load.  The 
point  at  which  the  counterbracing  is  to  begin  will  depend 
upon  the  numerical  data  for  each  case,  but  it  is  to  be  found 
by  the  preceding  method. 

The  moving  load  stress  in  any  web  member,  T^,  of  Fig.  4, 
between  the  end  A  and  the  centre  panel,  is  given  at  once 
by  Eq.  (18): 

(7;)=  +«(--+ ^)tc.^..r.^"    .     .     .     (39«\ 


TRUSSES    WITH    UNIFORM  PANEL   LOADS.  20m 

Or  between  the  end  B  and  the  centre  panel : 

(T;)  =  +  n  \^-^-)  w^jsec  a"      .     .     .     (39). 
Or  for  the  centre  web  member  corresponding  to  T^ : 

■{Tm)  =  +  m  i^L——^  li'j-sec  a    .     .     .     (40). 

Similarly,  Eq.  (19)  will  give  the  post  stresses  between  the 
end  A  and  the  centre  panel : 

{P^  =  —  n  { — ^ —  \  w  Y  sec  a"   .     .     .     (41a). 

Or  between  the  end  B  and  the  centre  panel : 

{P^)^-n{^-^)zvtseca'      .     .     .     (41). 

And,  for  the  centre  panel  of  an  odd  number: 

{P^)  =  -  VI  (^^^)  IV  ^  sec  a     .     .     .     (42). 

In  case  of  an  even  number  of  panels  in  the  span,  Eqs.  (40) 
and  (42J  will  not  be  needed. 

The  greatest  chord  stresses  will  exist  with  the  moving  load 
over  the  entire  span,  and  they  can  be  at  once  written  from 
Eqs.  (34)  and  (35)  by  making  W  equal  to  zero  and  then 
writing  iv  for  ]\\. 

By  thus  adapting  Eq.  (34)  for  the  lower  chord : 

Vn 
(«')  =  -\-  w\  —  {N  —  n'  +  i)  {ta7i  a   +  tan  «") 

+  (^«'  -  -  -  l)  tan  a" (43). 

The  same  process  of  adaptation  applied  to  Eq.  (35)  will 
give: 

(Un)  —  —  w  -  {N—  n  +  i)  {tan  a  +  tan  a")      .     .     .     (44). 


20;/  SPECIAL   NO/V-CONTINUOUS    TA'USSES. 

Again,  applying  the  same  process  to  Eq.  (38) : 

(m)  =  +  w  — ^^-5 '  {tan  a   +  tan  a  )    .     (45). 

o 

Eq.  (45)  will  give  the  stress  in  the  centre  panel  of  the 
lower  chord  when  the  number  of  panels  in  the  span  is  odd, 
but  that  equation  will  not  be  needed  when  the  number  is 
even. 

These  complete  all  the  expressions  for  stresses  required 
for  this  truss  used  as  a  through  structure. 

Case  V. —  The   Truss  of  Fig.  4  used  as  a  Deck  Structure, 

Fixed  Loads. 

The  formulae  of  Case  III.  are  to  be  applied  for  the  stresses 
in  this  case  precisely  as  they  stand,  with  the  notation  as  shown 
i?i  Fig.  4. 

Moving  Loads. 

When  the  moving  loads  traverse  the  upper  chord,  the 
notation  of  the  web  members  will  be  represented  by  Fig.  5, 
which  is  like  that  in  Fig.  4,  except  that  each  pair  of  braces 


^ 


Fig.  5. 


intersecting  in  the  lower,  or  unloaded,  chord  will  be  affected 
by  the  same  subscript.  The  remaining  notation  will  be  the 
same  as  that  used  in  connection  with  Fig.  4,  including  the 
uniform  moving  panel  load  w,  and  depth  of  truss,  d. 

Let  p  —  d  tan  (x"  =  p" , 
Let  p  —  d  tan  a    =  p' . 


7RUSSES    WITH    UNIFORM  PAN  EI   LOADS. 


200 


As  in  Case  IV.,  the  moving  load  will  be  taken  as  passing 
across  the  bridge  from  A  to  B.  The  shear  in  any  web  mem- 
ber, T,^,  between  A  and  the  centre  panel  of  an  odd  number  in 
the  lower  chord  will  be  the  same  as  that  indicated  in  £9.(39^:) 


diminished  by  7iu' 


In  other  words,  that  shear  is  the  saniv 


as  if  the  upper  chord  panel  points,  with  their  moving  loads, 

were  first  placed  over  the  corresponding  panel  points  in  the 

lower  chord  and  then  moved  back  to  their  actual  positions  ; 

in  the  first  of  these  positions  the  shear  would  be  indicated 

by  Eq.  (39^),  and  the  movement  back  to  the  actual  positions 

p" 
would  require  the  diminution  mv  y.     Hence,    the    stress    in 

any  Z,,  between  A  and  the  centre  panel  will  be : 


{T„)  =  +  ?i 


n  +  I 


+  niv 


P 


P  '  P'  . 

w  J  sec  a   —  nw  j-  sec  a 


1      2  7  j 


sec  a 


(46). 


The  stress  in   T^  [in  =  |  JV)  at  the  middle  panel,  corre- 
sponding to  Ti  in  Fig.  5,  will  be: 

{Tm)  =  +mw^^^   —^ "j]^   sec  a  .     .     (47). 


The  greatest  shear  in  the  P„^i  at  the  middle  panel,  corre- 
sponding to  /*5  in  Fig.  5,  will  exceed  that  in  T^n  by  i  zu. 
Hence  : 


{Pn.^l)  =  -'^ 


P  (m  +  I      p"\       I 


sec  a 


(48). 


Between  the  middle  panel  and  the  end  of  the  span  B,  with 
an  odd  number  of  panels  in  the  span,  T„^^x^  corresponding  to 
7^5  in  Fig.  5,  will  have  the  same  shear  as  {Prn  +  \)\  hence: 


(  7;„  + 1)  =  +  w 


p        m  +  I 
/      V     2 


I 

+  - 
2 


sec  a"     (49). 


20/ 


SPECIAL  NON-CONTINUOUS    TRUSSES. 


For  the  members  T^  with  subscripts  greater  than  {in  +  i), 
i.e.,  where  n'>{m  +  i) : 

(r„)  =  +  w  [^  j  ;«  (^— ^ ^  j  +  («  -  ?«  -  I) 


c^ 


+  ;;/  +  2       /' 


+ 


j^"^  a 


.-.(7;)=  +  w 


4  1  — '^ ^  —  {in  +   i)  —  in i— 

/  2  ^  / 


-("-■)|V7] 


sec  a 


■    (5o> 


With  an  even  number  of  panels  in  the  lower  chord,  Eqs. 
(47),  (48)  and  (49)  will  not  be  needed.  In  that  case  the  P^ 
and  T„,  {in  —  ^)  N  +  i) )  which  intersect  the  lower  chord  at 
its  centre  panel  point  will  have  the  same  shear  given  by 
Eqs.  (46)  and  (47).     Hence  : 


(51). 
(52). 


/  -r  \         ,  p  \  m  +  I        p" 

{Tra)  =  +  mw  ^  -|  — ^ ^  ]^sec  a 

/r>  \  p  \  in  +  I        p"  I  „ 


For  values  of  «  >  »« : 


fm  +  I       /"\        ,               fn  +  m-\-  I       p'\~] 
m  [—^  -r_^^in-  m)  (^ ^ ^JJ  sec  a 

~n{n  +  i)  /"  -/'  /n  „       ,     , 

— ^^ '  -  ni n  ^—   \sec  a'      (53), 

L        2  /  pA  ^^^' 


.•.(r„)=  + 


The  greatest  shear  in  any  /^,„  for  which  n  <  \  N,  or  less 
than  \{N  -\-  i),  according   as  the   number  of  panels   in   the 


TKUSSES    WITH    UNIFORM  PANEL   LOADS. 


20q 


span  is  odd  or  even,  respcctivel)-,  i.e.,  between  the  end  A  and 
the  centre  panel,  will  be  that  indicated  by  Eq.  (46).     Hence : 


/  (  «  +  I       /"  ) 


^    ^  /    (        2  /    ) 


sec  a 


(54)- 


Similarly,  when  the  number  of  panels  in  the  span  is  odd, 
and  where  «  >  (w  +  i),  Eq.  (50)  gives: 


i^n) 


W 


' p  \  n  {11  +  i) 
.7  '(         2 


P"  —  p'  P'  )        i~1 

—  {m  4-  i)  —  m         .       —  {n  —  i)  -  }  +  -  \  sec  a'  .     (55). 


The  end-post  stress  is 


{hP)  = zu  sec  01 


(56). 


EQ-   (55)    ^^''1   give    Eq.    (56)   by   making  n  =  N+  i    and 

N 


m  = 


When  the  number  of  panels  in  the  span  is  even  and  n  >  ni, 
Eq-  (53)  gives: 

(P.)  =  _  -/  [i(^)  - « ^'  -  „  t^ ,,,  „. .  (57). 


The  end-post  stress  is 


iV+  I 
{EP)  =  —  z^  sec  a' 


(58). 


Eq.  (57)  .will  give  Eq.  (58)  by  making  n  =  JV  +  i  and 
m  =  i(iV+  i). 

These  expressions  complete  all  that  are  required  for  the 
movinof  load  stresses. 


20r 


SPECIAL   NON-CONTINUOUS    TRUSSES. 


Case    VI. —  Truss  with  all  Web  Members  Equally  Inclined  to 
a  Vertical  Line — Warren  Girder. 

I. —  Through   Span — Fixed  Load. 

This  truss  is  illustrated  by  Fig.  6,  in  which  the  notation  is 
precisely  like  that  in  Fig.  4. 


The  inclination  of  all  web  members  to  a  vertical  line  =  a. 

/  =  length  of  span,  and/  —  panel  length. 

d=  depth  of  truss,  and  N  = i. 

P 

tan  ar  =  — , ;  and  sec  a  =  '^  \  -\-  tan^. 
2  d 

W'  =  uniform  upper  chord  panel,  fixed  load. 

Wi  =  uniform  lower  chord  panel,  fixed  load. 

The  reaction  at  either  end  of  the  span  due  to  the  fixed 
load  alone  is  given  by  Eq.  (26) : 


TV"  N  +  I 

R,  =  fw,+'^W   ....    (59). 


^^■y" 


The  stress  in  any  tensile  web  member  Tn,  between  that  end 
o(  the  truss  for  which  n  is  the  greatest  and  the  centre  of  the 
span,  is  given  by  Eq.  (29)  by  making  a"  =  a,  whether  the 
numb.^  of  panels  in  the  span  is  even  or  odd,  or  whether 
the  %uss  is  deck  or  through.     Hence : 


{T„)  =  + 


-^^ 


W' 


(IV'  +  W,)--^}  sec  a     (60). 


T/? LOSSES    WITH    UNIFORM  PANEL   LOADS.  20S 

The  equally  general  value  of  (/*„)  is  given  by  a  similar 
change  in  Eq,  (30).     Hence: 

W  =  -\^{n-^{W'  +W,)+^\seca  ,      (61). 

If  n'  represents  the  number  of  any  lower  chord  panel,  then 
Eq.  (34)  will  give  the  expression  for  the  stress  in  it  by  making 
tan  a  =  tan  a"  =  tan  a.     Hence  : 

(«')  =  +  [(«'  -  I)  {N-  n'  +  I)  {W  +  W^ 

^\{N{W^^  W)    +  W')-]  tan  a     .     .     ,     (62). 

In  precisely  the  same  manner  Eq.  (36)  will  give,  for  the 
upper  chord  stresses: 

{(7,,)=  -n{N-n  +  i){W'  +  W,)  tana     .     (63). 

The  end-post  stress  {£P)  will  always  be  found  by  making 
n  =  N"m  Eq.  (61),  which  will  then  be  reduced  to : 

{EP)  =-f^{W,+  W')+  ^)  sec  a  =  R,seca    .     (64). 


//. —  Through    Span — Moving  Load. 

Fig.  6  will  again  be  used,  and  the  uniform  panel  moving 
load  w  will  be  taken  as  moving  from  A  to  B. 

The  stress  in  any  member  T^  will  be  at  once  given  by  Eq. 
(18): 

(r„)=  +n{^-^)w^^seca  .  .  .  (65). 
Eq.  (19)  will  then  give  the  stress  in  any  post  (/*„) : 

(P„)=  -«(^)z^|..rr..  .  .  (66). 
Counterbracing  must  begin  when  the  moving  load  stress  in 


20/ 


SPECIAL   NON-CONTINUOUS    TRUSSES. 


Tn  or  P^  becomes  equal  to  that  of  an  opposite  kind  produced 
by  the  fixed  load,  as  explained  in  Case  IV. 

The  end-post  stress  {EP)  is   found  by  placing  n  =^  N  in 
Eq.  (66) : 


N 
{EP)  =  —  —  zv  sec  a 


i^7) 


The  stress  in  any  lower  chord  panel  ;/'  can  at  once  be  writ- 
ten by  making  W  =  o  in  Eq.  (62),  and  then  writing  w  for  W^ : 


(«')  =  + 


(;/'  -  i){N  -  n'  +  i)  + 


N- 


w  tan  a 


(68). 


Precisely  the  same  operation  applied  to  Eq.  (63)  will  give 
any  upper  chord  stress  U^  : 

{Un)  =  —  n  {N  —  n  +  \)w  ta^i  a      .     .     (69). 

Eq.  (69)  completes  the  expressions  for  the  stresses  in  all 
the  members  of  the  truss. 

Case  Mil.— Case  VI.  as  a  Deck  Span. 

I. — Deck  Span — Fixed  Load. 

The  formulae  for  web  and  chord  stresses  remain  exactly  as 
found  in  Case  VL,  I.,  ivitk  the  notation  as  given  in  Fig.  6. 

II. — Deck  span — Moving  Load. 

When  the  moving  load  passes  along  the  upper  chord  from 
A  to  B,  the  notation  of  the  web  members  must  be  as  shown 


Fig.  7. 


in  Fig.  7.  It  will  be  observed  that  the  only  change  from  Fig. 
6  consists  in  giving  the  same  subscript  to  any  two  web  mem- 
bers which  intersect  in  the  lower  chord. 


TRUSSES    WITH   UNIFORM  PANEL   LOADS. 


2011 


The  required  formulae  for  the  web  stresses  can  be  at  once 

written  from  those  of  Case  V.,  in  which  p'  —  p"  =  )^ p.     As 

usual,  the  uniform  panel  moving  load  will  be  w.     By  making 

P" 

•V  =  i  in  Eq.  (46): 


Similarly,  Eq.  (54)  gives  : 


n^       p 
+   —  w  -.  sec  a  . 
2         / 


«-       p 

—  w  ~  sec  oc  . 

2        / 


(70). 


(71). 


The  end-post  stress  {^EP)  will  result  from  Eq.(7i)  by  mak- 
ing ft  =  N  +  I  : 


tV  4-  I 
{hF)  =  —  — w  sec  a 


.     (72). 


Any  lower  chord  stress  {ti)  can  be  written  from  Eq.  (62)  by 
making  W^  =  o  and  then  writing  w  for  ]\"  : 

[n)  —  [{n  —  \){N  —  n   ^  1)+  ^  (A^  -r-  i)]  w  tan  a  .    (73). 

In  the  same  manner  Eq.  (63)  gives,  for  any  upper  chord 
stress : 

[Un)  =  —  n  {N  —  n  +  i)  u>  tan  a  .     .     .     (74). 

All  formulae  for  the  deck  truss  are  thus  made  complete. 
Case  VIII. —  Vertical  Ties  and  Inclined  Posts. — Hoive   Truss. 

I. —  Through  Span — Fixed  Loads. 

The  truss  to  be  treated  in  this  case  is  shown  in  Fig.  8,  in 
which  all  inclined  web  members  are  in  compression.     If  the 

a      6 c  d 

\     '  \  T  '-' 


Fig. 


compression  members  and  chords  are  built  of  timber,  it  is  the 
common  Howe  truss.  The  notation  will  be  that  already 
used. 


20Z/  SPECIAL  NON-CONTINUOUS    TRUSSES. 

W  =  uniform  upper  chord  panel  fixed  load. 
W^  =  uniform  lower  chord  panel  fixed  load. 
/  =  length  of  span.    /  =  panel  length. 


d  =  depth  of  truss.     Tan  a  —  ^.     Sec  a  =  y  i  ^-cL 
N=L      u 

P 

The  posts  P.y  to  /'j,  inclusive,  in  Fig.  8,  are  counterbraces, 
and  will  not  be  stressed  by  the  fixed  load  ;  hence  they  are  to 
be  neglected  in  finding  the  fixed  load  stresses. 

The  reaction  at  each  end  of  the  truss  due  to  the  fixed  load 
only,  will  be : 

Rr=-^i}V  ^-W,)      ....    (75). 

It  is  agreeable  to  the  notation  of  the  figure  to  consider  the 
main  web  members  in  the  right  half  of  the  truss.  The  shear 
in  any  inclined  web  member  P^  between  the  centre  panel  and 
the  end  B  of  the  span  will  be  the  reaction  at  A^  less  the 
loads  n  (W  +  W^),  exactly  as  indicated  by  Eq.  (3)  of  Case  I. 
Hence: 

{P„)=  -  {ri  -  j^  {W  +  W,)  sec  a       ....     (76). 
Similarly : 

(zj- +  [(«-^)([^'4- fro- w^]  .  .   .   {77)' 

In  Eqs.  {76)  and  {77),  n  '^( —  -f  i  j  for  an  odd  number  of 

panels  in  the  span;  and  n  >  ( ■  )    for  an  even  number 

of  panels.     In  the  latter  case,  however,  the  fixed  load  stress 

1  •   1    11  •        1       .        •         T  A^  +     I 

m  the  middle  vertical  tie  is,  it  ?/?  = : 

2 

{Tm)=    +    W, (78). 


tr'usses  with  uniform  panel  loads.         2(ym 
The  end-post  stress  is  found  by  making  «  =  iV  in  Eq.  (76): 

{EP)  =  -~{W'  -^  IV,)  sec  a       .     .     (79). 

The  lower  chord  stresses  will  be  found  by  taking  mo- 
ments about  the  upper  chord  panel  points.  If  n'  is  the  num- 
ber of  any  lower  chord  panel  between  the  centre  and  end 
panels,  inclusive,  the  equation  of  moments  will  be: 

(«')  d=-{lV'  +  IV,)  .  71' p  -  n' {IV  +  W,).  ^^-^^p 

...     [n'^^^  ^'l^i^N-n'  ^  i){W'  +  W,)tana       .      (80). 

which  is  identical  with  Eq.  (11). 

The  upper  chord  stresses  at  once  follow: 


(d)  =-(i)    1 

{c)  =  -  (3) 
etc.=  —  («') 


1^ (81). 


A  vertical  end-post  at  each  end  of  the  span  will  carry  sim- 
ply the  load  at  its  top,  and  none  of  the  preceding  expressions 
will  be  changed. 

//. —  Through   Span — Aloving  Load. 

The  uniform  panel  moving  load  will  be  w. 

The  greatest  moving  load  stress  in  any  counterbrace  (/*„), 

for  which  n  <  —  for  an  odd  number  of  panels  in  the  span,  or 

N  —  I 
n  <_ for  an  even  number  of  panels,  will  be  exactly  sim- 
ilar to  that  given  by  Eq.  (18) : 

(/*„)=  —  n  (^ j  zv^.  sec  a     .     .     .     (82). 


20X 


SPECIAL   NON-CONTINUOUS    TRUSSES. 


If  n  =  N, 


{P„)  =  {EP)  =   -  —  w  sec  ex  .  (82«). 


Eq.  (82)  also  gives  the  stress  in  any  main  post  P^  as  it 
stands,  by  giving  n  the  proper  value,  which  will  be  greater 
than  the  limits  indicated  above. 

Any  main  member  Z'„  will  have  the  stress: 

(7;)=  +n(^^)^^^~i        ....     (83). 

The  lower  chord  stresses  can  be  written  at  once  from  Eq. 
(80): 

{n')  =  -I {N  —  n'  ■{-  i)w  tan  a      .     .     (84). 


As  before : 


W=-(3)    f 
etc.=  -  («')  J 


(85). 


Expressions  for  all  the  stresses  are  thus  completed. 


///. — Deck  Span — Fixed  Load — ]\[o7'ing  Load. 

Fig.  9  shows  this  truss  as  a  deck  structure.     The  notation 
is  like  that  shown  in  Fig.  8,  except  that  the  subscripts  of  the 


vertical  braces  {T^  are  so  written  as  to  make  each  pair  of 
web  members  intersecting  in  the  lower  chord  take  the  same 
subscript. 


TRUSSES    WITH   UNIFORM  PANEL   LOADS.  20/ 


Fixed  Load. 

The  expressions  for  all  the  stresses  due  to  the  fixed  load 
are  precisely  the  same  as  those  in  Case  VIII.,  I.,  Eqs.  (75) 
to  (81),  inclusive,  but  the  subscripts  for  the  vertical  members 
Tn  must  be  used,  as  shown  in  Fig.  8. 

Moving  Load. 

All  the  chord  stresses  due  to  the  moving  load  will  be  given 
by  Eqs.  (84)  and  (85)  exactly  as  they  stand,  as  the  notation 
for  the  chords  is  the  same  in  Figs.  8  and  9. 

All  the  web  stresses  due  to  the  moving  load  will  be  given 
by  Eqs.  (82),  (82^),  and  (83)  just  as  they  stand,  but  the  iveb 
notation  shown  in  Fig.  9  must  be  used. 

Vertical  end-posts  will  change  nothing,  as  they  will  carry 
that  load  only  which  rests  at  their  tops. 

Tables  for  Computations. 

The  following  Tables,  which  explain  themselves,  will  enable 
the  preceding  equations  or  formulae  to  be  applied  to  actual 
cases  with  considerable  economy  of  labor. 

The  numerical  value  of  the  expression  {n  —  i )  {N  —  ;?'  +  i) 
can  readily  be  taken  from  the  first  of  the  Tables  in  the 
following  manner.  Clearly  [ji  —  i)  (.V  —  n  -\-  \)  —{ri  —  i) 
{N  —  («'  —  i)  —  I  +  i)  =  {n  —  i)  [A^—  {n  —  i)].  Hence,  if 
{n  —  i)  is  placed  for  n  in  the  Table,  the  desired  result  will 
be  the  tabular  result  diminished  by  {n'  —  i).  As  an  ex- 
ample, let  iV  =  17  and  ;/  =  7.  Then  take  ;/  =  ;/'—  i  =  6. 
The  tabular  result  is  72,  which,  diminished  by  6,  is  72—6 
—  66  =  in'  —  i)  {N  —  n'  +  iV  Indeed,  the  desired  result  is 
always  found  in  the  Table  in  the  column  with  heading 
{N  —  I)  and  opposite  n  =  («'  —  i). 


202 


SPECIAL   NON-CONTINUOUS    TRUSSES. 


Values  of  n  {N—  ti  +  i). 


n 
I. .. 

3... 
4... 
5--- 
6... 

3 

3 
4 
3 

4 
4 

6 
6 
4 

5 

5 
8 

g 

8 

5 

6 

6 
10 
12 
12 
10 

6 

7 
7 

12 
15 
16 
15 
12 

7 

8 

8 
14 
18 
20 
20 
18 
14 

8 

9 

9 
16 
21 
24 

25 

24 
21 

16 

10 

10 
18 
24 
28 
30 
30 
28 

24 
18 
10 

II 

II 

20 
27 
32 
35 
36 
35 
32 
27 
20 
II 

12 

12 
22 
30 
36 
40 
42 
42 
40 
36 
30 
22 
12 

13 

13 
24 
33 
40 
45 
48 
49 
48 
45 
40 

33 
24 
13 

14 

M 
26 
36 
44 
50 
54 
56 
56 
54 
50 

36 
26 
14 

15 

'5 
28 
39 
48 
55 
60 
63 
64 
63 
60 
55 
48 
39 
•28 
15 

16 

16 
30 
42 

52 
60 

66 
70 
72 
72 
70 
66 
60 
52 
42 
30 
16 

17 

17 
32 
45 
S6 
65 
72 
77 
80 
81 
80 

77 
72 

65 
56 
45 
32 
17 

18 

18 

48 
60 

7? 
78 
84 
88 
90 
90 
88 
84 
78 
70 
60 
48 
34 
18 

19 

36 
51 
64 

8^^ 
91 
c6 
99 

ICO 

99 
96 

9> 
84 
75 
f4 
51 
36 
'9 

20 

20 
38 
54 
68 
80 
90 
98 
104 

Toa 

8 

10. . . 

no 

Tn8 

104 
98 
90 
8n 

i5--- 

16 

68 

18 

::::  .;;; 

54 

38 

j 

1    j 

1 



Values  0 


r  /«     +     I   \. 


n  = 

I 
I 

2 

3 

3 
6 

4 
10 

5 

15 

6 

21 

7 
28 

8 
36 

9 

45 

10 

55 

II 

66 

12 

78 

13 

9' 

14 
105 

15 

120 

16 

17 

18 

171 

19 

20 

«  («  +  i) 
2 

136 

153 

190 

210 

Values  0 


■H^y 


n  = 

I 
0 

2 
I 

3 
3 

4 
6 

5 
10 

6 
15 

7 
21 

8 
28 

9  10 
36  45 

II 

55 

12 
66 

13 
78 

14 

15 
105 

16 

17 
136 

18 

19 

20 

«  («  —  i) 

91 

120 

153 

171 

190 

2 

Art.   7. — Position    of   Moving    Load    for   Greatest   Shear   and   Greatest 

Bending. 

That  method  of  computation  which  treats  the  moving  load 
as  composed  of  a  system  of  isolated  weights  requires  some 
simple  method  of  finding  the  greatest  possible  shear  in  a 
given  panel  for  a  given  system  of  loading.  Among  the  first 
to  use  such  a  method  was  Theodore  Cooper,  C.  E.  ;  and  the 
results  of  the  following  investigation  are  the  same  as  those 
determined  by  him. 

The  method  and  formula  first  developed  apply  to  any 
single  system  of  triangulation  so  far  as  the  web  stresses  are 
concerned,  but  for  the  chord  stresses  they  only  apply  to  such 
a  system  when  composed  of  alternately  vertical  and  inclined 
members.  Subsequent  modifications  for  web  members  all 
inclined  will  be  made  for  the  chord  stresses. 


GREATEST  STRESSES. 

Case  I. 


21 


Let  the  moving  load  consist  of  the  advancing  weights  Wy, 
IVi,  W3,  ....  IV,^  separated  by  the  distances  a,  b, 
c,  d,  etc. ;  then  let  the  weights  W^i,  W^,     ....      Wn'  be 


Fig.  I. 


found  in  the  panel  DC,  in  which  it  is  desired  to  find  the 
greatest  possible  shear,  and  at  the  distances  b' ,  b" ,  .  . 
from  C ;  it  being  understood  that  the  moving  load  advances 
from  F  toward  G.  The  last  load  W^  is  found  at  the  distance 
X  from  F.  The  length  of  span  FG  is  /,  while  the  length  of 
panel  Z>C  is/.  With  the  assumed  position  of  loading,  the 
reaction  R  dit  G  will  be : 


R  =  \ 


W, 


a  +  b  +  c  + 


+   W. 


+   W, 


b  +  c  +  d  -{- 


c  +  d  ^ 


+  ^„f 


+  X 


+   X 


+  ;r    .     .     .     (I) 


The  parts  of  the  weights  W^,  W^, 


.     .     resting  on 


22  SPECIAL  NON-CONTINUOUS    TRUSSES. 

h'  h" 

DC,   which  pass  to  D,  are  W^   -,  W^  —,    .    .     .     .     Hence 
the  shear  in  the  panel  DC  will  be 

S  =  R-(W,^  +  W,^'+ )    ....     (2) 

If  the  train  advance  by  the  amount  A  x,  the  new  reaction 
R',  at  G,  takes  the  value : 

R'  =  R  +  {IV,+  IV,+  W,+   .     .     .     .+W,)^;.     (3) 

and  the  new  shear  will  become  : 

S'=--R'-{m^-+W,~+  .   .   .   )-{W^+W,+  .  .  .)—    (4) 
V  p  '  P 

-{W,+  W,+  .   .   .   )^ (5) 

Or,  S'  -S  =  ~\^{W,+  W^^-  W,+  .   .    .    +  W^) 

-{W,+  W,.  ..){/-=  n)]^ (6) 

Whenever  S'  —  S  becomes  equal  to  zero,  S'vvill  be  either  a 
maximum  or  a  minimum.  If  the  difference  is  positive  just 
before  it  becomes  equal  to  zero,  S'  will  be  a  maximum,  and 
that  is  the  only  case  of  interest  in  the  present  connection. 
Hence,  by  placing  5'  —  5  equal  to  zero,  the  following  condi- 
tion is  obtained  : 

n{W,  +  W.,  +  .  .  .)=W,+  W,+  W,  +  .  .  .  +  Wn  ..{7) 

The  shear  in  the  panel  in  question  will  therefore  take  its 
greatest  value  when  n  times  the  moving  load  which  it  contains 
is  equal  to,  or  most  nearly  equal  to,  the  entire  moving  load  on 
the  bridsre. 


GREATEST   STRESSES.  23 

That  equality  will  seldom  or  never  exist  unless  one  of  the 
weights  W  is  placed  on  a  panel  point,  since  IVi  +  IV2  +  .  .  . 
is  seldom  or  never  an  exact  divisor  of  the  entire  load  on  the 
bridge.  If  a  weight  rests  on  a  panel  point,  any  part  of  such 
a  weight  may  be  taken  as  acting  in  one  adjacent  panel  and  the 
remainder  in  the  other  ;  the  desired  equality  may  thus  be 
obtained. 

In  case  Eq.  (7)  should  hold,  the  position  of  the  moving 
load  is  a  matter  of  indifference  so  long  as  the  panel  in  ques- 
tion contains  the  same  number  of  loads  JVi  +  IV^  +  ■  •  •> 
as  there  is  no  trace  of  d',  d",  etc.,  in  that  equation.  A  load 
may  then  always  be  taken  as  resting  at  the  rear  extremity  of 
the  panel  where  the  greatest  shear  in  it  exists. 

These  considerations  frequently  essentially  simplify  com- 
putations. 

When  the  value  of  x  has  been  found  for  the  position  of 
the  greatest  shear,  the  latter  being  determined  by  the  preced- 
ing method,  Eq.  (2)  may  be  put  in  the  following  convenient 
shape  by  the  aid  of  Eq.  (i)  : 

S='^[IV^  a +{Wi+W,)d +  {IV,+W,  +  W,)c+     .     .     .     . 

P 

-f     .     .     .     +{IV,  +  W,  +  .     .     .     +IV„._,)?]     .     (8) 

The  sign  (?)  stands  for  the  distance  between  the  wheel  con- 
centrations Wn'^i  and  Wn',  since  the  latter  rests  directly  at 
the  panel  point  in  question. 

It  is  thus  seen  that  all  the  parts  of  Eq.  (8)  may  be  taken  at 
once  from  tables,  except  that  term  involving  x. 

Case  II. 

In  the  preceding  case  it  has  been  supposed  that  for  the 
greatest  shear  in  DC,  the  front  weight  W^  is  found  between 
/?  and  C;  but  let  JV^  +  IV2  +  etc.,  be  supposed  between  D 
and  G. 


24  SPECIAL   NON-CONTINUOUS    TRUSSES. 

With  the  notation  remaining  the  same  as  before,  the  shear 
^  will  become: 

s  =  R-{w,  +  w,  +  Qtc.)-{wA  +  w,j+  .   .   .);(9). 

while  S'  takes  the  value  : 

-{W,^W,+     .     .    .     )~. 

Hence  for  a  maximum,  the  following  expression  must  never 
become  negative : 

-{W,  +  W,+     .    .     .     )(^-^  =  n^^=o.    .    .(10). 

But  Eq.  (lo)  is  identical  with  Eq.  (7).  Hence,  the  same 
conditions  for  a  maximum  obtain  wherever  may  be  the  head 
of  the  moving  load.  The  second  member  of  Eq.  (8),  how- 
ever, must  contain  the  negative  sum  of  all  the  weights  be- 
tween D  and  G. 

EXAMPLE. 

If  each  one  of  the  weights  W^,  IV^,  etc.,  is  equal  to  any- 
other,  i.e.,  if  they  are  all  uniform,  and  if  a  —  b  —  c  —  d 
=  .  .  .  =  p,  Eq.  (7)  shows  that  the  front  weight  W^ 
must  be  taken  at  the  first  extremity  of  the  panel  in  question. 
The  same  result  holds  if  the  first  weight  VVi  is  not  exceeded 
in  amount  by  any  that  follows  it,  provided  that  a,  b,  c,  etc., 
still  equal  /. 

In  cases  where  the  same  system  of  concentrated  loads  is  to 
be  used  for  a  number  of  spans,  it  will  be  shown  that  the 
tabulation  of  the  products  of  the  sums  of  the  weights  Wi,  W^, 
etc.,  by  the  distances  a,  b,  c,  etc.,  can  be  advantageously  used 


GREATEST  STRESSES. 


25 


to  shorten  and  simplify  computation.  In  other  cases,  how- 
•ever,  the  quickest  and  simplest  method  is  partly  graphical ; 
it  is  as  follows : 


( 

■% 

F 

^ 

^ 

3 

B 

' 

f 

"^^^^^A 

1 

\ 

VI 

W 
Fig.  2. 

1 

In  Fig.  2  let  AB  be  the  length  of  span,  and  W  any  weight 
resting  anywhere  in  the  span.  Erect  a  vertical  at  B  and  let 
BC  represent  W  by  any  convenient  scale ;  then  draw  the 
straight  line^C  The  vertical  intercept  WD  w'lW  represent 
the  reaction  at  B  due  to  W,  by  the 
same  scale  on  which  BC  represents 
that  weight. 

In  the  same  manner  if  5/^  repre- 
sents W^,  then  W^E  will  represent 
the  reaction  at  B  due  to  W^.  Thus 
there  must  be  as  many  verticals 
BF,  BC,  etc.,  as  there  are  different 
weights  resting  in  the  span,  and  the 

total  reaction  at  B,  for  any  given  position  of  the  moving 
load  will  be  the  sum  of  the  vertical  intercepts  WD,  W^E, 
etc.,  erected  at  each  load  ]V  for  that  position. 

h'  h" 

The  negative  shears  Wx  -,   W^  —,  etc.,  appearing  in  Eq.  (2) 

are  most  readily  found  in  the  same  manner.  If  GH,  Fig.  3, 
is  a  panel  length  and  [Fany  weight  represented  by  the  verti- 
cal line  GK,  drawn  to  any  convenient  scale,  while  WH  is 
equal  to  b' ,  b" ,  etc.,  then  the  vertical  intercept  WE  between 


G"// and  A'// will  represent  W 
-It  G  due  to  W, 


,   /k— ,  etc.,  I.e.,  the  reaction 


26  SPECIAL   NON-COi\ TINUO U S    TRUSSES. 

If  the  reaction  at  B,  Fig.  2,  is  then  given  by  ^IVB,  and 

/'  A" 

W\  — \-lV« 1-  etc.  =  ^  WE,  Fig.  3,  the  shear  (see  Eq.  (2)  ) 

/  '  P 

will  be : 

S=2WD-  2W£. 

If  Figs.  2  and  3  are  drawn  on  profile,  or  cross-section  paper, 
the  shears  for  any  span  can  be  found  with  great  ease  and 
rapidity. 

Position  of  Moving  Load  for  Greatest  Bending  Mojuent. 

The  Fig.  and  notation  of  the  preceding  cases  will  be  used 
in  connection  with  this.  Moments  will  be  taken  about  the 
panel  point  C,  horizontally  distant  /'  from  G.  There  will 
be  supposed  to  be  n  weights  in  front  of  C  {i.  e.,  between 
C  and  G),  and  the  weight  W^„.  will  be  taken  at  the  distance 
x'  from  C  towards  G.  The  bending  moment  M  will  then 
take  the  value : 


M=Rl'  - 


Wi(a  +  I?  +  c  +     .     .     .      +  x) 
+  W^{        b  +  c  +     .     .     .     -\-x') 

+  W,,x'. 


Or,  after  taking  the  value  of  R  from  the  preceding  cases: 

M=j[W,a  +  {W^^  W,\  d  +  (W,  +  M^,  +  Ws)e+     .     .     . 

+  {w,  +  iv,  +  w,+  .  .  .  ^iVn)x]-w,a-{w^  +  jj:;)d 

-{lV,  +  1V.,  +  W,)e-    .    .    .     -{W,+  W^  +  W^+    ... 
^  W^.)x' (Ill 

If  the  train  advances  by  the  amount  Ax,  the  moment  be- 
comes : 

M'  =  M^^-j{W,  +  W^  +  W^+    .     .     .    ^W^)  Ax-{W,  +  W^ 
+-...+  W,,)  AX, (12). 


GREATEST   STRESSES.  27 

Hence,  for  a  maximum,  the  following  value  must  never  be 
negative : 

-f     .    '.     .      +W„.)\=o (13). 

Or,  the  desired  condition  for  a  maximum  takes  the  form  : 


I  ~  W,+W^  +  W^+     .     .     .     -\-l^V„ 


(H). 


It  will  seldom  or  never  occur  that  this  ratio  will  exactly 
exist  if  Wn'  is  supposed  to  be  a  w/io/e  weight  ;  hence,  IV,,.  will 
usually  be  that  part  of  a  whole  weight  at  C  which  is  neces- 
sary to  be  taken  in  order  that  the  equality  (14)  may  hold. 

It  is  to  be  observed  that  if  the  moving  load  is  very  irregu- 
lar, so  that  there  is  great  and  arbitrary  diversity  among  the 
weights  W,  there  may  be  a  number  of  positions  of  the  train 
which  will  fulfil  Eq.  (14),  some  one  of  which  will  give  a  value 
greater  than  any  other  ;  this  is  the  absolute  maximum  desired. 

Since  IV^'  will  always  rest  at  a  panel  point  for  the  greatest 
bending  moment,  x'  in  Eq.  (11)  may  always  be  put  equal  to 
zero  when  that  equation  expresses  the  greatest  value  of  the 
moment.     The  latter  then  becomes: 

M  =  j[lV,a  +  {IV,+W,)d  +     .    .    .     +  {W,+W,+     .    .    . 

+  W^)x]-lV,a-{U\+JV,)b-     .     .     .     -iU\+lV, 
+  ..     .     .     +^„.-i)? (15)- 

In  this  equation,  of  course  x  corresponds  to  the  position  of 
maximum  bending,  while  the  sign  (?)  represents  the  distance 
between  the  wheel  concentrations  r^,j_i  and  JV„. 

It  is  known  that  for  any  given  condition  of  loading  the 
greatest  bending  moment  in  the  beam  or  truss,  will  occur  at 
that  section  for  which  the  shear  is  zero.  But  if  the  shear  is 
zero  at  that  section,  the  reaction  R  must  be  equal  to  the  sum 
of  the  weights  {li\  +  /f  2  +  ^^  3  +  •  •  •  +  ^^«)  between  G 
and  C;  the  latter  now  being  that  section  at  which  the  greatest 


28  SPECIAL   NON-CONTINUOUS    TRUSSES. 

moment  in  the  span  exists.     Hence  for  that  section  Eq.  (14). 
will  take  the  form  : 

/'  R 


I  ~  w^  +  w.  +  w,+  .  .  .   +  w,: 

or,  the  centre  of  gravity  of  the  load  is  at  the  same  distance 
from  one  end  of  the  truss  as  the  section  or  point  of  greatest 
bending  is  from  the  other.  In  other  words,  the  distance  between 
the  point  of  greatest  betiding  for  any  given  system  of  loading, 
and  the  centre  of  gravity  of  that  loading  is  bisected  by  the  cen- 
tre of  span. 

If  the  load  is  uniform,  therefore,  it  must  cover  the  whole 
span. 

It  will  be  observed  that  Eq.  (15)  is  composed  of  the  sums 
of  Wi,  Wi  +  W^,  etc.,  multiplied  by  the  distances  a,  b,  c,  etc., 
precisely  as  in  Eq.  (8),  hence  the  same  tabulation  as  there 
indicated  may  be  used  to  advantage. 

Limitations  of  the  preceding  methods. 

The  preceding  methods  are  limited  to  a  single  system  of 
triangulation.  By  the  use  of  certain  assumptions  in  refer- 
ence to  the  distribution  of  the  loading  between  two  or  more 
systems  of  triangulations  in  the  same  truss,  a  somewhat  simi- 
lar investigation  might  be  made  for  such  cases,  but  such 
analysis  would  not  be  rigorously  exact.  Hence  it  is  as  well 
to  pursue  the  usual  method  and  assume  that  each  system 
acts  as  an  independent  truss,  then  place  the  moving  load  in 
such  a  position  for  each  system  that  the  front  panel  load  for 
that  system  will  be  the  greatest  possible.  This  panel  con- 
centration will  then  be  the  forward  panel-moving  load,  and 
the  succeeding  ones  may  either  be  those  concentrations 
which  actually  correspond  to  the  forward  one,  or  may  be 
supposed  to  be  composed  of  a  uniform  load  equivalent  to 
the  concentrated  one.  The  latter  plan  will  be  employed 
hereafter. 

Application  of  the  preceding  method  to  an  all-inclined  tveb 
system. 

As  was  observed  at  the  beginning  of  this  Art.  the  analy- 
sis for  chord  stresses,  as  already  given,  is  directly  applicable 


GREATEST   STRESSES.  29 

to  a  single  system  of  triangulation  in  which  a  vertical  web 
member  is  found  in  each  panel.  The  general  demonstration, 
however,  is  easy. 

If  the  web  members  are  all  inclined,  the  formulae,  as  al- 
ready given,  are  directly  applicable  in  any  case  to  the  deter- 
mination  of  stress  in  that  chord  which  does  not  carry  the 
moving  load,  since  moments  are  taken  about  the  panel  points 
of  the  chord  traversed  by  the  moving  load. 

But  let  it  be  required  to  determine  the  position  of  the 
moving  load  for  the  maximum  stress  in  DC,  Fig.  i,  and  the 
expression  for  the  corresponding  moment.  As  before,  let 
the  load  move  from  F  toward  G.  Let  q  represent  the  hori- 
zontal distance  of  D  from  H,  i.  e.,  q  =  KH ;  evidently  q  is 
constant  for  the  same  span.  Let  x^  represent  the  horizontal 
distance  from  H  of  the  first  load  to  the  left  of  D,  and  let  that 
load  be  represented  by  W^„.     That  portion  of  the  loads  rest- 

inij  in  DC,  which  is  transferred  to  D,  is  2W-.     I'  will  now 
f  '  p 

represent  GD  4-  q. 

By  taking  moments  about  H,  Eq.  (11)  will  take  the  form : 

M  ^j\W^a  -^  {\\\  -V  W^b  +  .  .  .  +(^Fi  +  IF2  +  .  .  .  +  W,^x-\ 
-  lV,a-{lV,+  W,)d  -...  -{IV,+  W,  +  ...  +  m„)x, 
~^2Wd' (16). 

By  advancing  the  train  Ax,  since  Ax  =  Ax^=  Ab\  Eq. 
(13)  will  become : 

M' -M=  AxY-^{W,^W.^  ...  +^„)-(I^,+  I^2+...+  ^n) 

-^:s^l=o (17). 


The  condition  for  a  maximum  or  minimum  then  takes  the 
shape : 


L  _  ^1  +  ^2  +  .  .  .  +  m^  +  f^w 


(18). 


30  SPECIAL  NON-COA'TINUOUS    TRUSSES. 

Eqs.  (i6)  and  (i8)  are  the  general  expressions  of  which 
Eqs.  (15)  and  (14)  are  special  forms. 

After  X  and,  hence,  x-^  have  been  determined  by  the  aid  of 
Eq.  (18),  the  maximum  moment  will  be  given  by  Eq.  (16),  in 
which  the  tabulations  already  indicated  can  be  advantage- 
ously employed. 

Application  of  preceding  methods  to  a  system  of  concentra- 
tions followed  by  a  uniform  load. 

If  the  uniform  load  does  not  reach  to  the  panel  under  con- 
sideration, which  is  usually  the  case,  W^  in  Eqs.  (7),  (14)  and 
(18)  represents  the  total  uniform  load  on  the  bridge,  but  the 
formulae  are  in  no  wise  changed.  In  Eqs.  (8),  (15)  and  (16), 
however,  it  is  to  be  observed  that  while  W^^  again  represents 
the  total  uniform  load,  x  will  represent  the  distance  from  its 
centre  of  gravity  to  the  end  of  the  span  {i.  e.,  half  the  length 
covered  by  the  uniform  load),  also  that  the  distance  between 
IVn  and  W^-i  will  be  equal  to  x  plus  the  space  which  sepa- 
rates Wn-x  from  the  front  of  the  uniform  load. 

In  the  case  of  the  existence  of  this  uniform  load  it  will 
happen  that  IVn^  will  not  rest  at  a  panel  point.  The  last  term 
in  the  negative  expressions  of  the  second  members  of  Eqs. 
(8)  and  (15)  will  then  h&  {W^  +  W^  +  .  .  .  +  IV„.)  x' ; 
x'  being  the  distance  of  W^'  in  front  of  the  panel  point  C. 
Eq.  (16)  is  general,  and  needs  no  change  on  this  account. 

If  the  concentrations  are  so  few  that  the  uniform  load  ex- 
tends over  a  portion  of  the  panel  in  question,  the  observa- 
tions made  above  still  hold.  But  in  addition  to  them,  ^„.  or 
W^n  will  represent  the  amount  of  uniform  load  in  the  panel, 
and  x  or  x^  will  represent  the  distance  from  its  centre  of 
gravity  to  the  panel  point.  The  interval  or  space  between 
Wn_x  or  ^n-1  and  W^'  or  W\  will  then  be  the  distance  from 
either  of  the  former  to  the  centre  of  gravity  of  the  uniform 
load. 

Finally,  in  Eq.  (17)  or  (18)  '^W  will  be  either  wholly  or 
partly  composed  of  uniform  load. 

Modifications  for  Skew  Spans. 
If  a  skew  bridge  is  under  treatment,  the  preceding  methods 


GREATEST   STRESSES. 


31 


apply  in  all  respects,  so  far  as  the  general  principles  are  con- 
cerned. 

It  will  be   sufficiently  accurate   in   all  cases    to    treat  the 


Fig.  4. 


Fig.  5. 


moving  load  as  if  it  were  passing  along  the  centre  line  LL 
L'L  of  each  track. 

In  the  case  of  the  single  track  bridge,  Fig.  4,  if  the  load  is 
passing  from  right  to  left,  the  moving  load  does  not  rest  on 
the  truss  CD  until  it  passes  the  point  C ,  and  continues  to  act 
on  that  truss  until  it  passes  to  the  same  distance  to  the  left 
of  D,  it  being  borne  in  mind  that  all  moving  load  is  trans- 
ferred to  the  trusses  by  transverse  floor  beams  placed  nor- 
mal to  the  axis  of  the  bridge.  It  results  from  these  con- 
siderations that  if  the  load  passes  from  right  to  left  in  Fig. 
4  and  along  LL,  the  reactions  at  D  will  be  greater  than 
a  half  of  those  at  K  by  the   amount  of  the  half  products 


32  SPECIAL   NON-CONTINUOUS    TRUSSES. 

of  the  total  load  corresponding  to  those  reactions  by  the 
ratio 

CC 
CD 

Hence,  if  R  is  any  reaction  at  K  and  '^W  the  total  load, 
the  corresponding  reaction  at  D  will  be : 

2      ^  CD        ' 

On  the  other  hand,  with  the  load  moving  in  the  same 
direction,  the  reaction  at  A  will  be  : 

2      -^CD 
In  Eqs.  (8),  (15),  and  (16),  then,  there  is  to  be  written  for 
{IV,  +  W,+     .     .     .     +lVn)x 
the  expression, 

{W,  +  W,+     .    .    .     +W,){x±CC\ 

according  to  the  direction  in  which  the  train  is  moving ;  dut 
the  negative  portions  are  to  remain  ujtchanged.  It  is  to  be 
remembered  that  the  quantity  x  is  to  be  measured  on  the 
centre  line. 

In  the  case  of  the  double  track  skew  bridge  of  Fig.  5,  in 
which  there  are  the  two  trusses  AB  and  CD  only,  precisely 
the  same  observations  hold.  For  one  track,  however,  CC"  is 
to  be  used  and  CC  for  the  other.  Separate  computations, 
are  to  be  made  for  each  track  for  each  truss. 

If  there  are  three  trusses  in  Fig.  5,  each  pair  of  trusses  con- 
stitutes a  single  track  bridge  for  the  track  between,  and  is  ta 
be  treated  precisely  as  Fig.  4. 

If  the  skew  is  so  great  that  one  or  more  floor  beams  have 
their  end  or  ends  resting  on  the  masonry,  obvious  modifica- 


GREATEST  STRESSES. 


33 


tions  must  be  made  according  to  the  preceding  general  prin- 
ciples. 

TJie  G  rap  J  deal  Method. 

With  convenient  means  for  constructing  an  accurate  equi- 
librium polygon  with  a  large  number  of  loads,  this  method  is 
a  very  rapid  one  for  either  shears  or  moments.  A  perfect 
familiarity  with  the  principles  and  operations  of  Art.  45  is 
here  supposed. 

Let  the  entire  moving  load  for  a  given  truss  be  represented 
by  the  system  of  forces  i,  2,  3,  4,  and  5,  in  Fig.  6.  They  are 
given  in  actual  position  under  the  polygon  PKO.     P  is  the 


B      5       r^ 


pole,  EF  the  load  line,  and  PO  an  indefinite  horizontal  line 
normal  to  EF.  As  usual,  the  moving  load  is  supposed  to 
move  from  O  toward  P.  The  polygon  PQK  ....  NO  is 
formed  in  the  ordinary  manner  by  making  its  sides  parallel 
to  the  lines  radiating  from  P.  For  reasons  that  will  pres- 
ently be  evident,  QF  should  be  continued  considerably 
beyond  C,  while  NO  should  be  carried  somewhat  below 
the  horizontal  line  through  P. 

After  the  constructions  indicated  have  been  made  to  any 
3 


34  SPECIAL  NON-CONTIA'UOUS    TRUSSES. 

convenient  scale,  let  the  span  under  consideration  be  laid 
off  to  the  same  scale  by  which  the  horizontal  separations 
between  the  loads,  i,  2,  3  and  4,  etc.,  are  laid  down,  and  let 
indefinite  vertical  lines  be  drawn  through  the  panel  points, 
but  let  all  this  latter  construction  be  made  on  tracing  cloth. 
In  order  to  avoid  confusion,  neither  the  panel  points  nor  the 
vertical  lines  through  them  will  be  shown  in  the  Fig. 

In  the  first  place,  let  the  position  of  the  moving  load  for 
the  maximum  shear  in  some  web  member  be  determined  by 
Eq.  (7;,  and  in  this  position  let  the  loads,  i,  2,  3  and  4,  be 
supposed  to  rest  on  the  truss  ;  and  let  B\  represent  the  dis- 
tance between  the  last  concentration  and  the  right  hand  of 
the  span  {j.  c,  B4  is  x  of  Eq.  (8)  ).  Now  let  the  tracing 
cloth  be  superimposed  on  the  equilibrium  polygon  in  such  a 
manner  that  AB  shall  represent  the  span  ;  then  erect  the 
verticals  BD  and  AC,  and  draw  CD,  to  the  latter  of  which 
PH  is  drawn  parallel.  FH  will  then  represent  the  reaction 
at  the  left  end  of  the  span,  /.  e.,  at  A.  If  from  this  reaction 
the  negative  shear  shown  in  Eq.  (8),  or  found  by  the  method 
of  Fig.  3  be  subtracted,  the  result  will  be  the  shear  desired 
in  the  web  member  under  consideration.  In  this  manner  all 
the  maximum  shears  may  be  found. 

Again,  let  it  be  required  to  find  the  greatest  moment  at  a 
given  panel  point,  for  which  the  moving  load  has  been  found 
by  Eqs.  (14)  or  (18),  to  occupy  such  a  position  that  the  dis- 
tance from  the  right  end  of  the  span  to  the  last  concentra- 
tion (/.  e.,  X  in  Eqs.  (15)  and  (16)  )  is  represented  by  B'  5  in 
Fig.  6.  Also  with  the  position  of  moving  load  thus  deter- 
mined, let  it  be  supposed  that  the  load  2  rests  at  the  panel 
point  considered.  Now,  let  the  tracing  cloth  be  so  super- 
imposed on  the  equilibrium  polygon  that  A' B  will  represent 
the  span,  then  erect  the  vertical  lines  B' D'  and  A'C\  and  join 
CD' .  Since  load  2  was  found  at  the  panel  point,  at  which 
the  moment  is  to  be  determined,  KL  will  represent  the  max- 
imum moment  in  question.  Each  linear  unit  in  KL,  meas- 
ured by  the  same  scale  to  which  the  span  and  distances,  t-2, 
2-3,  etc..  are  laid  down,  will  represent  as  many  moment  units 
as  there  are  force  units  in  the  pole  distance  PT.     If  PT  is 


GREATEST   STRESSES.  35 

in  pounds  and  KL  in  feet,  the  FT  X  KL  will  be  the  mo- 
ment desired  in  foot-pounds.  In  the  same  manner  all  maxi- 
mum moments  may  be  determined.  In  every  position  of 
moving  load  required,  the  vertical  intercept  between  the 
closing  line  and  equilibrium  polygon,  drawn  through  the 
panel  point  considered,  will  represent  the  maximum  moment 
at  that  point.  PH'  drawn  parallel  to  CD'  will  give  FH'  as 
the  reaction  at  F,  though  it  is  of  no  special  value  in  this  con- 
nection. The  application  of  this  method  to  the  different 
panels  of  a  truss  will  give  all  the  greatest  chord  stresses. 

This  method  is,  of  course,  subject  to  all  the  modifications 
that  have  been  outlined  for  the  various  special  cases  and 
conditions.  It  cannot,  however,  be  applied  to  the  mov'.ig 
load  on  skew  bridges.  The  reaction  for  the  centre  line  of 
the  track  must  be  reduced  to  the  trusses,  while  the  negative 
moments  of  the  loads  in  advance  of  the  panel  point  which 
serves  as  the  moment  origin  remain  unchanged.  The  gener- 
ality of  this  method  is  not,  therefore,  comple;te. 

The  Maximum  Floor-bcairi  Reaction. 

The  moving  load  iscarried  to  ea;ch  transverse  floor  beam  by 
the  adjacent  stringers.  Hence  each  floor  beam  is  a  pier  for 
two  adjacent  spans  of  string-trs,  and  it  becomes  necessary  to 
determine  that  position  of  the  moving  load  on  those  two 
spans  which  will  subjec'c  the  floor  beam  to  its  greatest 
load. 

In  the  Fig.  let  a  sect.ion  of  the  beam  be  shown  at  R,  while 
/  and  /'  are  the  two  adjacent  stringer  spans  traversed  by  the 
moving  load  ;  then  let  the  ^s  be  measured  from  the  right 
and  left  ends  of  /,  and  /',  while  W,  IV\  etc,  Jl\.  W„_,  etc.,  rep- 
resent the  weigh  ts  or  wheel  concentrations  resting  in  the  two 
spans,    the  reaction  R  will  then  have  the  value: 

IV,  xi  +  W^x.-,  -h  etc,      Wx  -^   VV^  x'  -t-  etc,        .     . 
R^. ^_1_. + ^- .    (19). 

If  the  whole  system  of  loading  move  to  the  left  by  the 
distance  ^x,  th  e  new  reaction  will  be  : 


36 


SPECIAL   NON-CONTINUOUS    TRUSSES. 


(JFi  +    W^  +  etc.)    A,r        (^F+    ^i  +  etc.)  A;»r 
^   ~  ^  I,  ^  I 

In   that  position  which  gives  a   maximum    or   minimum,, 
J^i  —  R  =  0;  hence : 


{Wi  +   JV2  +   ^s  +  etc.)  -  ^{IV+  W  +  W"  +  etc.)  (20). 
It  will  seldom  happen  that  Eq.  (20)  will  be  satisfied  unless 


^5-■ 


-->U-- 


|<- 1 av — >i 


k- 


h 


^ & 


x--^ 


_l 


r 

L. 


W,  W2        etc.       ""  etc. 


w 


w 


J 


L 


nr 


_IL 


Fig.  7. 


a  concentration  rest  on  the  poii?t  R,  so  that  the  proper  por- 
tion of  it  may  be  taken  for  one  span  or  the  other,  precisely 
as  in  the  problems  of  maximum  shear  and  maximum  mo- 
ments. 

Ordinarily  the  two  adjacent  spans  afe  equal,  or 


/=/, 


W,  +  W,  +  etc.  =  W  +  W'  +  etc.   .    .    (21). 


Eq.  (21)  shows  that  when  the  two  span.'5  are  equal,  the 
amounts  of  load  each  side  of  R  must  also  be  eUual. 

After  the  proper  position  of  loading  has  bee^n  determined^ 
Eq.  (19)  will  give  the  maximum  reaction  desirec^* 

Article  8.— Fixed  Weight.  » 

As  the  weight  of  a  structure  forms  a  very  considerable 
portion  of  its  total  load,  it  becomes  a  matter -of  importance 

/ 


FIXED    WEIGHT.  3/ 

to  find  at  least  an  approximate  value  for  it,  when  the  moving 
load  has  been  once  assumed. 

The  weight  of  ties,  guard  timbers  or  rails,  rails,  spikes,  etc., 
may  be  taken  at  350  to  475  pounds  per  lineal  foot  of  single 
track,  and  forms  an  invariable  part  of  the  fixed  weight,  (/.  r., 
weight  of  structure),  since  it  is  independent  of  span,  length 
of  panel,  or  depth  of  truss.  With  ordinarily  heavy  traffic 
and  standard  gauge,  400  pounds  per  lineal  foot  is  usually 
taken. 

After  having  fixed  the  weight  of  track,  the  stringers  are  to 
be  designed.  A  very  little  experience  will  enable  their 
weight  per  lineal  foot  to  be  assigned  in  advance,  so  that  the 
total  load  resting  upon  them  may  be  used  in  making  compu- 
tations. These  track  stringers  are  almost  invariably  plate 
girders,  and  the  main  object  of  the  computation  is  to  deter- 
mine the  area  of  flange  section.  If  that  area,  as  computed, 
makes  the  weight  of  the  stringer  very  different  from  that 
assumed,  it  will  be  necessary  to  again  assume  a  weight, 
guided  by  the  results  of  the  first  computation,  and  re-calcu- 
late for  the  flange  area  ;  and  then  repeat  the  operation  until 
sufficient  accuracy  is  obtained. 

The  weight  of  the  track  stringer  thus  obtained  becomes  a 
part  of  the  load  sustained  by  the  floor  beam.  The  weight 
per  lineal  foot  of  the  latter  is  then  to  be  assumed,  and  calcu- 
lations made  and  repeated,  if  necessary,  precisely  as  in  the 
•case  of  the  stringers.  With  a  very  little  practice  the  weights 
of  the  stringers  and  floor  beams  may  be  so  accurately 
assigned  in  advance,  that  a  re-calculation  is  seldom  or  never 
necessary. 

The  lateral  and  transverse  systems  of  bracing  should  next 
be  subject  to  computation,  and  as  the  wind  pressure  is  their 
only  load,  their  own  weight  does  not  affect  the  operations  ; 
hence  no  re-calculation  will  ever  be  required. 

The  remaining  calculations  are  those  of  the  truss  proper, 
and  the  accurate  assignment  of  their  own  weights  presents 
more  difficulty  than  any  other  part  of  the  operation.  Ex- 
perience, however,  enables  even  this  weight  to  be  quite 
closely  taken  in  advance.     For  each  single  track  of  standard 


38  SPECIAL   KON-CONriNUOUS    TRUSSES. 

gauge  the  weight  of  two  through  trusses,  designed  for  the 
moving  load  taken  in  Art.  9,  together  with  the  lateral  sys- 
tem, in  pounds  per  lineal  foot,  may  be  approximately  taken 
at  five  times  the  length  of  the  span  in  feet,  or  double  that 
amount  for  a  two  truss  double  track  structure.  If  this  is  not 
sufficiently  close,  one  or  more  re-calculations  will  be  required. 
For  spans  over  two  hundred  and  fifty  feet  in  length,  this  rule 
gives  too  small  results. 

By  thus  designing  the  floor  system,  and  lateral  and  trans- 
verse bracing  before  the  computations  are  made  for  the 
trusses,  it  is  only  necessary  to  assign  before  each  step,  the 
weight  of  that  part  immediately  under  consideration,  and, 
hence,  enables  the  actual  weight  to  agree  very  closely  with 
the  assumed,  without  re-calculation. 

If  the  bridge  is  of  the  deck  variet}'  and  carries  the  ties  di- 
rectly on  the  upper  chord,  so  that  they  act  as  a  transverse 
load  on  the  latter,  thus  obviating  a  system  of  track  stringers 
and  floor  beams,  the  total  fixed  weight  will  be  reduced  about 
125  pounds  per  lineal  foot.  If,  on  the  other  hand,  such  a 
bridge  carries  a  regular  system  of  stringers  and  floor  beams, 
the  truss  weight  may  be  the  same  as  if  it  were  a  through 
bridge. 

If  the  span  is  less  than  about  eighty  or  more  than  about  two 
hundred  and  fifty  feet,  the  weight  per  lineal  foot  will  some- 
what exceed  the  value  given  by  the  preceding  rule. 

If  the  bridge  is  a  highway  structure,  the  same  general 
method  of  operations  (and  taken  in  the  same  order)  is  to  be 
followed  as  for  a  railway  bridge.  A  rule  for  truss  weights  can- 
not, however,  be  so  easily  given,  because  the  moving  load  is 
so  very  variable  ;  they  may  equal  or  exceed  those  of  railway 
structures  of  the  same  span,  or  may  not  exceed  a  third  of  that 
value.  The  weight  of  a  highway  floor,  if  of  plank  and  timber 
joists,  will  be  from  twenty  to  twenty-five  pounds  per  square 
foot. 

Art.  9 — Single  System  of  Bracing  with  Two  Inclinations. 

The  first  case  taken  will  be  that  shown  in  Fig.  2  of  PI.  I. 
The  span  j  is  120  feet;  depth  d  20  feet  ;  panel  length  /  20 


RAILWAY  BRIDGES.  39* 

feet;  angle  NAR,  i8°  30',  and  angle  NAM,  33°  40'.  The 
moving  load  will  be  taken  as  two  coupled  consolidation  loco- 
motives, each  with  weights  distributed  as  shown  in  Fig.  i, 
Art.  "]"],  followed  by  a  uniform  load  of  1.5  tons  per  lineal  foot. 
The  bridge  will  be  supposed  to  be  a  single  track  "  through  " 
structure.  Each  truss  will  be  taken  to  weigh  approximately 
240  pounds  per  lineal  foot.  The  upper  lateral  bracing  will  be 
taken  at  thirty  pounds  per  lineal  foot  for  each  truss.  The 
total  weight  concentrated  in  each  of  the  upper  chord  panel 
points  will  then  be  10  x  270  =  2,700  pounds.  The  fixed 
weight  concentrated  in  the  lower  panel  points  will  be  taken 
at  9,300  pounds.  Hence,  if  W  is  the  upper  chord  fixed  panel 
load  and   W  the  same  for  the  lower : 

W  =  2700  pounds  =  1.35  tons,  for  one  truss. 
W  ■=  9300       "         =  4.65     " 
tan  NAR  —  0.333  t^n  NAM  -  0.666 

sec       "       =   1.054  sec       "         —   1.202 

In  the  above  fixed  weight,  the  ties,  rails,  guard  timbers,  etc., 
were  taken  at  400  pounds  per  lineal  foot. 

The  stresses  due  to  the  fixed  load  only,  in  each  of  the  truss 
members  will  first  be  found,  and  it  will  be  convenient  to  begin 
by  determining  those  in  the  web  members. 

On  page  7  it  is  shown  that  the  stress  in  any  web  member 
of  a  truss  with  horizontal  chords  is  the  vertical  shear,  multi- 
plied by  the  secant  of  its  inclination  to  a  vertical  line.  But 
the  vertical  shear  in  any  web  member  of  such  a  truss  is  simply 
the  algebraic  sum  of  all  the  vertical  forces  or  weights  {includ- 
ing the  end  reaction^  between  the  end  and  the  zveb  Member  in 
question. 

In  web  member  5,  for  example,  the  fixed  load  shear  will  be 
the  difference  between  the  reaction  R  and  the  weights  at  A, 
B,  M  and  Z;  or,  since  the  truss  is  symmetrical  with  the 
centre,  the  shear  will  be  the  weight  at  C  added  to  half  the 
weight  at  K.  In  fact,  as  a  general  principle,  when  the  truss 
and  its  load  are  symmetrical  with  the  centre,  the  shear  in  any 
web  member  will  be  the  load  between  that  member  and  the 


40 


SPECIAL  NON-CONTINUOUS    TRUSSES. 


(I). 


centre.     Hence,  the  shear  in  the  web  member  of  half  the  truss 
will  possess  the  following  values  : 

Shear  in  6  =    i^  =    2.325  tons 

'<    5  =     1 J/F  +     W  =    3.675     " 

"   4  =  illV  +     W  =^    8.325     " 

''  ;^  =  l^W  +  2W'  =  9.675  '' 
"  "  2  =  2hJ'V  +  2IV'  =  14.325  " 
"         "     I   =  2hW  +   3IV'  =   15.675      " 

If  the  plus  sign  indicates  tension  and  the  minus  sign  com- 
pression, the  web  stresses  will  take  the  following  values : 

Stress  in  6  =  +     2.325   x  1.202  =  -f     2.79  tons 

"  "   4  =  +     8.325   X  "     =  +   10.00     " 

"  "    2  =  +  14.325   X  "     =  +   17.19     " 

"  "    5  =  -    3-675   X  I-054  =  -     3-87     " 

a  u      2    =z    —       9.675     X  "        =    —     10.  18        " 

"  "      I     =    —    15.675     X  "        z=    —     16.52        " 


^       (2). 


On  the  same  page,  7,  it  was  shown  that  the  increment  of 
chord  stress  at  any  panel  point  is  equal  to  the  algebraic  sum 
of  the  horizontal  components  of  the  web  stresses  intersecting 
at  that  point ;  /;?//  t/iosr  horizontal  components  are  the  vertical 
shears  multiplied  by  the  tangents  of  the  respective  inclinations 
to  a  vertical  line.      Hence  : 


The  chord  increment  at : 

A  =  15.675    X   ^  +  14.325   X  I 

B  =     9.675   X   ^  +  8.325   X  I 

^  =     3-675   X   ^  +  2.325   X  I 


14.79  ^^^^ 

8.79 

2-79 
5-24 


^(3)- 


R  =  15.675  X   1  +  o 

3/=  9.675  X   i  +  14.325    X  I  -  12.79 

Z  =  3.675  X   1  +  8.325   X  I  =  6.79 

K=        o  +  2.325  X  I  =  1.55 

Hence  the  following  are  the  upper  chord  stresses : 
(i)     =  -  14.79  =  -  14-79  tons      ) 

(2)  =  -  (14.79  +  8-79)  =  -  23-58     "        [   .     (4)- 

(3)  =  -  (23-58  +  2.79)  =  -  26.37     "        ' 


The  lower  chord  stresses  will  take  the  values : 


J?  AIL  IV AY  BRIDGES. 


41 


(i)     =  5.24   tons 

(2)  =     5.24  +  12.79  =  18.03     " 

(3)  =  18.03  +     6.79  =  24.72     " 


(5). 


When  the  position  of  the  moving  load  is  once  determined  for 
a  required  maximum  stress,  the  latter  is  found  from  the  re-ac- 
tion and  panel  loads  by  precisely  the  same  general  methods 
used  with  the  fixed  loads.  Hence  the  proper  position  of  the 
moving  load  is  first  to  be  found  for  each  of  the  web  stresses. 

Eq.  (7)  of  Art.  7  is  an  expression  of  the  condition  which 
obtains  with  the  greatest  shear  in  any  web  member.  The 
number  of  panels,  n,  is  6.  W^,  W»,  IV^,  etc.,  are  the  single 
locomotive  weights  given  in  Art.  yy ;  IV^  has  the  value  3.75 
tons ;    W2,  W3,  Wi  and  W  5,  6  tons,  etc.,  etc. 

If  the  moving  load  passes  on  the  bridge  from  the  right,  and 
W^  rests  at  the  foot  of  web  member  10,  U\  .  .  .  \V^  will 
be  on  the  bridge,  and  )i  times  3.75  {i.e. ,6  x  3.75  =  22.50  tons) 
will  be  less  than  IVi  +  IV^  +  .  .  .  +  IV^  =  27.75  tons.  But 
^  ^  (3-75  +  6)  tons  is  much  greater  than  the  total  moving 
load  on  the  bridge.  Hence,  Ifo  at  G  is  the  position  desired 
for  web  member  10 ;  W^  will  then  be  8.083  ^^et  from  G  toward 
H.  At  this  point  the  tabulation  mentioned  in  connection 
with  Eqs.  (8)  and  (15)  of  Art.  7,  may  be  used. 

The  tabulation  for  the  two  locomotives  is  P'iven  herewith. 


I 

2 

3 

4 

I 

15,000  X    8.o3 

121,200 

121,200 

2 

39,000  X    5.75 

224,250 

345.450 

3 

63,000  X    4.50 

283,500 

628,950 

4 

87,000  X    4.50 

391,500 

1,020,450 

5 

111,000  X    7.08 

785,880 

1,806,330 

6 

126,000  X    4.83 

608,580 

2,414,910 

7 

141,000   X    5.67 

799.470 

3,214,380 

S 

156,000   X    4.83 

753.480 

3,967,860 

9 

171,000  X    9.00 

1,539,000 

5,506,860 

10 

186,000  X    8.o3 

1,502,880 

7,009,740 

II 

210,000  X    5.75 

1,207,500 

8,217,240 

12 

234,000  X    4.50 

1,053,000 

9,270,240 

13 

258,000  X    4.50 

1,161,000 

10,431,240 

14 

282.000  X    7.08 

1,996,560 

12,427,800 

15 

297,000  X    4.83 

1.434.510 

13,862,310 

16 

312,000   X    5.67 

1,769,040 

15.631.350 

17 

327,000   X    4.83 

1,579.410 

17,210,760 

18 

342,000  X    4.00 

1,368,000 

18,578,760 

42  SPECIAL    NON-CONTINUOUS    TRUSSES. 

A  little  consideration  of  the  table  will  make  its  composition 
evident.  It  is  reproduced  from  a  table  in  actual  use,  and  is 
given  in  pounds  and  foot  pounds. 

When  [F2  rests  at  G,  the  x  of  Eq.  (8)  will  be  5.25  feet. 

It  should  be  explained  that  each  quantity  in  column  4  is 
the  sum  of  all  the  preceding  and  opposite  numbers  in  column 
3.  As  an  example,  5,506,860  is  the  sum  of  all  the  numbers 
in  column  3  from  the  top  down  to  and  including  the  ninth. 
Column  4,  then,  represents  the  positive  parenthesis  in  Eq. 
(8)  of  Art.  7,  less  the  term  multiplied  by  x. 

When  W^  rests  at  G",  the  x  of  Eq.  (8),  Art.  7,  will  be  5.25 
feet.  Hence  that  equation  in  connection  with  the  preceding 
table  gives  : 

f,,  I     /i  1 1000  i02045o\        I        121200 

Shear  in  10  =  ( x  5.25  ^ =^^^  ) x . 

I20\2  2/20  2 

=  3650  pounds  =  1.825  tons (6). 

This  quantity  will  shortly  be  needed  again. 

If  the  same  locomotive  weight  W.^  rest  at  H,  W^     .     .     . 

W^,  or  one  complete  locomotive,  will  be  found  on  the  bridge, 
and  n  times  (3.75  +  6)  or6  x  9.75  =  58.50  tons  is  still  in  excess 
of  42.75  tons,  t.  c,  half  the  entire  locomotive  weight.     Hence, 

W^  at  H  is  the  position  of  moving  load,  which  gives  the 
greatest  shear  to  web  member  8,  and  x,  in  Eq.  (8)  of  Art.  "jy 
becomes  2.83  feet.     That  equation  then  gives  : 

„,         .    „         I      /171000         „        3q6786o\        I        121200 

Shear  in  '^  —      -     ^ x  2.83  +  -^-^ —  —  x . 

120   \      2  2/20  2 

=  15520  pounds  =  7.76  tons (7). 

If  the  same  weight,  W.^,  be  placed  at  K,  it  will  be  found  that 
W-Q,  will  rest  at  the  right  extremity  of  the  span,  there  will 

therefore  be  eleven  weights  on  the  bridge.  Since  6  x  (1200a 
+  7500)  =  1 17000  >  105000,  this  position  of  the  moving  load 

gives  the  greatest  shear  in  brace  6.     Eq.  (8)  of  Art.  7  thea 

gives,  since  x  =  5.75  feet : 

„,         .    -       8217240        I         121200 

Shear  m  o  = -—^ x  . 

2  X  120       20  2 

=  31400  pounds  =  15.7  tons (8). 


RAIL  IV A  V  BRIDGES.  43 

If  W-i,  be  placed  at  Z,  it  will  be  found  that  six  times 
{W^  +  W.^  is  less  than  W^  .  .  .  .  W^r,,  which  will  then 
rest  on  the  bridge ;  but  6  ( W^  -v  W^  ^  W^  >  {W^  .  .  .  . 
H^ie),  hence  W^  must  rest  at  L  for  the  greatest  shear  in  4. 
The  value  of  x  will  then  be  4.83  feet.     Hence  : 

I    A 1 2000         „         i38623io\        I        345450 

S/iear  m  4  =  (  ^ x  483  +     ^    ^^      j  -  —   x  ^^^^ 

^I20\2  2/20  2 

=  55480  pounds  =  27.74  tons (9). 

In  the  case  of  brace  2,  the  condition  of  maximum  shear 
obtains  with  JFg  at  M.  This  position  of  moving  load  places 
10.5  feet  of  the  uniform  load  of  1.5  tons  per  lineal  foot  on 
the  bridge.  The  value  of  x  in  Eq.  (8)  of  Art.  7,  conse- 
quently, becomes  5.25  feet.  The  tabulation  given  above 
is  not  quite  sufficient  to  completely  cover  this  case ;  although 
the  result  in  line  18  forms  by  far  the  greater  portion  of  the 
moment  product  which  must  be  divided  by  the  span,  in 
order  to  obtain  the  shear.  The  application  of  Eq.  (8)  of  Art. 
7,  to  this  particular  case  becomes,  then  : 


I      / 342000  3000 

—   [^ X   10.5    +  ^^ — 

120    \       2  •'2 

i857876o\       345450 


Sheaf  tn  2  =  — ^  y^ —      ^   iO-5    +  "^T    ^   ^°-5   ^  5-^5    +- 


2   X  20 

=  84430  pounds  =  42.2  tons (10). 

The  moving  load  shears  in  web  members  i,  3,  5,  7  and  9, 
will  be  precisely  the  same  as  those  in  2,  4,  6,  8  and  10  re- 
spectively, because  each  pair,  as  10  and  9,  8  and  7,  etc.,  inter- 
sect in  a  chord  which  carries  no  moving  load.  The  web 
stresses  due  to  the  moving  load  will  then  take  the  following 
values : 

Stress  in  10  =  +  1.825  x  1.054  =  ~     1.92  tons.  - 

"       "  8  =  +  7.760  X      *'      =  +     8.1 

"       "  6  =  +     15.7  X  1.202=  -^   18.87    "        r  •  (lO- 

"       "  4  =  +  27.74  X       "      =  +  33.35     " 

*       "  2  =  +  42.20  X       ''      =  +  50.72     •' 


44  SPECIAL  NON-CONTINUOUS    TRUSSES. 

Stress  in    9  =  —  1.825  x  1.202  =  —  2.19    tons.  1 
u       <.     7=  —  7.762  X       "     =  —  9.33      "       I 

"    "  5  = -157  X  1.054  =  -16.49  "     r  •  (^2). 

"       "     3  =  —  27.74  X       "      =—29.13     "        i 
"       "      I  =  —42.20  X       "      =—44.30     "       j 

By  comparing  (11)  and  (2),  it  is  seen  that  the  moving  load 
stress  in  web  member  (10)  is  of  a  kind  opposite  to  that  caused 
by  the  fixed  load  in  web  member  3,  while  those  two  members 
are,  in  reality,  identical  ;  the  latter  is  compression  and  the 
former  tension.  Since  it  is  evident  that  no  piece  of  ma- 
terial can  be  compressed  and  extended  at  the  same  time,  it 
is  clear  that  the  resultant  stress  will  be  the  numerical  differ- 
ence or  algebraic  sum  of  the  induced  stresses.  In  other 
words  : 

1/  the  action  of  forces  external  to  a  piece  of  material  tend  to 
subject  that  portion  of  material  to  stresses  of  opposite  kinds,  the 
resiiltant  stress  will  be  equal  to  the  numerical  differejice  of  the 
opposite  stresses,  and  will  be  of  the  same  kind  as  the  greater. 

By  comparing  (11)  and  (12)  v/ith  (2),  it  will  be  seen  that  in 
all  the  web  members  of  that  half  of  the  truss  first  traversed  by 
the  train,  the  fixed  and  moving  load  produce  stresses  of  opposite 
kinds.  The  moving  load  stresses  predominate  in  the  mem- 
bers near  the  centre  of  the  span,  but  near  the  ends  the  fixed 
load  stresses  are  the  greatest.  Ir.  web  member  8,  the  resul- 
tant stress  is  8. 1  —  3.87  =  4.23  tons  of  tension  ;  but  if  the 
bridge  carries  no  moving  load,  it  is  subjected  to  3.87  tons  of 
compression.  Now,  that  member  may  be  so  designed  and 
constructed  that  it  can  resist  tension  or  compression,  accord- 
ing to  the  demands  upon  it ;  in  such  a  case  it  is  said  to  be  co2in- 
terbraced.  If,  however,  it  is  formed  to  resist  compression  only, 
the  member  14  must  be  introduced  between  /Tand  (7  to  take 
the  moving  load  shear  in  tension.  In  order  to  provide  for 
the  movement  of  the  load  in  the  opposite  direction,  13  must 
be  introduced  between  L  and  D.  Such  web  members  as  13 
and  14  are  called  counterbraccs.  Other  web  members  than 
counterbraces  are  called  main  braces  or  main  web  members. 
The  duty  of  counterbraces,  then,  or  of    count erbraced  main 


RAILWAY  BRIDGES.  45 

web  members,  is  to  transfer  moving  load  shear  to  the  farther 
abutment  or  pier. 

It  is  now  clear  that  the  extent  to  which  a  main  web  mem- 
ber must  be  counterbraced  is  found  by  taking  the  excess  of 
the  moving  load  stress  over  that  caused  by  the  fixed  load. 
At  first  sight  it  would  appear  that  the  same  method  should 
hold  in  determining  counterbrace  stresses  ;  since  it  may  be 
supposed  that  the  main  brace  will  carry  shear  until  its  fixed 
load  stress  is  neutralized.  This  presupposes,  however,  that 
there  is  such  an  exact  adjustment  of  members  that  each  will 
perform  just  the  amount  of  duty  assigned  to  it.  As  that  is 
an  end  that  can  never  be  confidently  realized,  it  is  only  pru- 
dent to  suppose  that  the  counterbrace  takes  all  the  moving  load 
shear,  and  this  will  be  assumed  in  all  that  follows.  This 
procedure  appears  the  more  advisable  when  one  reflects 
that  counterbraces  are  subject  to  the  greatest  fatigue  of  all 
truss  members,  and  that  the  amount  of  metal  concerned  is 
trifling. 

It  now  becomes  necessary  to  determine  where  the  coun- 
terbraces are  to  begin.  Since  the  fixed  and  moving  load 
stresses,  or  shears,  neutralize  each  other  in  equal  amounts,  it 
at  once  results  that  counterbraces  or  counterbraced  zveb  mem- 
bers must  begin  at  that  point,  or  with  that  web  member,  in 
which  the  stress  or  shear  produced  by  the  moving  load  is 
greater  than  that  of  the  opposite  kind  produced  by  the  fixed  load 

"  Stress  "  or  "  shear  "  is  used  indifferently,  as  the  former  is 
simply  the  product  of  the  latter  by  the  secant  of  the  inclina- 
tion to  a  vertical. 

In  order,  therefore,  to  find  the  stress  in  a  counterbrace,  it 
is  only  necessary  to  ascertain  the  moving  load  shear,  and 
multiply  it  by  the  secant  of  the  inclination.  The  secant  of 
the  angle  between  counterbrace  13  or  14,  and  a  vertical  line 
is  1.944 ;  and  since  the  shear  it  has  to  carry  is  by  Eq.  (i  i)  'j.'j6 
tons: 

Stress  in  counterbrace {i/^)='j.y62  x  1.944=  + 1 5.09  tons.(i 3^ 

Again,  by  comparing  (11)  with  (2)  it  is  seen  that  the  fixed 
load  shear  in  10  (or  3)  is  —  9.675  tons,  while  the  moving  load 


46  SPECIAL    NON-CONTINUOUS    TRUSSES. 

shear  is  +1.825  tons.  Hence  13  and  14  are  the  only  counter, 
braces  needed. 

It  is  farther  seen  that  in  the  half  of  the  truss  traversed  last 
in  order  by  the  train,  the  stresses  produced  by  the  fixed  and 
moving  loads  are  the  same  in  kind.  Hence,  in  the  main 
braces  the  fixed  and  moving  loads  induce  stresses  of  the  same 
kind  and  the  resultant  is  si^nply  the  numerical  sum. 

A  tabulation  of  all  the  resultant  web  stresses,  then,  gives 
the  following  values : 


Web 

mem 

ber  I  =  —60.82  tons.  ~ 

n 

" 

3  =-39-31     " 

t< 

<t 

5  =  -  20.36     " 

« 

(< 

2  =  +  67.91     " 

(( 

<< 

4  =  +43-35     " 

« 

<( 

6  =  -f  21.66     " 

« 

a 

14  =  +  15.09     " 

(14). 


The  moving  load  chord  stresses  remain  to  be  found,  and  it 
will  be  necessary  to  resort  to  the  methods  of  Art.  7  in  order 
to  determine  the  proper  positions  for  the  stresses  in  the  vari- 
ous panels. 

Since  none  of  the  web  members  are  vertical,  the  positions 
of  moving  load  for  the  greatest  stresses  in  the  lower  chord 
panels  will  be  given  by  Eq.  (18)  of  Art.  7.  For  this  case,  q 
in  that  equation  will  be  one-third  the  panel  length. 

In  finding  the  stresses  in  lower  chord  panels  2  and  3,  it  is 
necessary  to  bring  10.4  and  7  feet,  respectively,  of  the  uni- 
form load  on  the  bridge.  The  condition  of  greatest  stress  in 
the  lower  chord  end  panel  coincides  with  that  for  brace  i, 
and  it  will  only  be  necessary  to  multiply  the  greatest  shear 
in  that  brace  (already  determined)  by  the  tangent  of  its  in- 
clination to  a  vertical  line. 

In  determining  the  greatest  stresses  in  the  upper  chord, 
moments  are  taken  about  the  lower  chord  points,  and  Eq. 
(14)  of  Art.  7  will  be  used. 


RAIL  WA  Y  BRIDGES. 


47 


The  application  of   Eqs.  (i8)  and  (14)  of  Art.   7  give  the 
following  results  : 

Lower  did  2  .  .  j  =  ^. .  W\    =  IV^  . .  4r,=6.67  ft  . .  2  x=  10.4  ft. 

"  =  i    .  IV„._^=  W^ 2  ^=  10.4  " 

"  =  i  .  .     "      =  ^5 2  ,r=   6.4  " 

"    =  /r,! 2  ^=19.0  " 


t( 

"    3•• 

upper 

"    I  . . 

<i 

"    2  . . 

(< 

"3.. 

U     1 


Eq.  (16)  of  Art.  7,  applied  to  the  lower  chord,  gives  by  the 
aid  of  the  tabulation  already  employed  the  moments: 

In  lower  chord  2  ;       M  =  ^-^\  — '^  ■    ^' —  +  (2  x  171000+  10.4 

X  1500)  5.2    —  i^^^^ ^ —  X  6.67  —  1(12000  X  15.5 

+  1 2000  X  1 1  +  7500  X  3.92), 

.-.     J/=  1,936,400  ft.  lbs. 
In  lower  chord  3  ;      M  —  ^\  -  4-  (2  x    17 1000  4-  7 


X  1500)3.5 


1,806,330       126000 


X  T.2  —  \  (7500  X  15.67 


+  2x7500x7.58), 

.-.     J/=  2,658,490  ft.  lbs. 

Eq.  (15)  gives  for  the  upper  chord  moments: 

18,578,760 


In  upper  chord   i  ;  M  =■  \ 


+  (2  X  171000  +  10.4 


X  1500)  5.2 


345450 , 
2 


J/=  1,685,425  ft.  lbs. 


48  SPECIAL  NON-CONTJNUOUS   TRUSSES. 

"18,578,760 


In  upper  chord  2  ;      M 


+  (  2  X    171000  +  6.4 


X  1500)3.2 


1,806,330 


In  upper  chord  3  ;      M  =  ^ 
X  1500)  9.5  — 7500X  122.5 


J/ =2,568,335  ft.  lbs. 

18,578,760 


2 


+  (2  X  171000  +  19 


/8,2i7,240  ^      \ 

-  [2         -  7500 X 62.5 j 

M=  2,305,315  ft.  lbs. 


The  negative  moments  (  — 7500X  122.5)  ^"d  (—7500x62.5) 
occur  in  the  last  moment  above,  for  the  reason  that  IVi  is 
found  2.5  feet  off  the  span  to  the  left  when  upper  chord  3 
takes  the  maximum  bending. 

It  is  notable  that  the  moment  in  upper  chord  panel  3  is 
less  than  that  in  panel  2,  while  it  is  yet  more  important  to- 
observe  that  the  uniform  load  of  1.5  tons  per  lineal  foot  gives 
the  greatest  bending  moment  in  the  centre  panel  of  the  upper 
chord.  A  panel  uniform  load  is  10x1.5  =  15  tons  =  3O0OO 
pounds,  and  the  reaction  with  this  load  on  the  whole  bridge 
is  2.5  X  15  =  37.5  tons=750OO  pounds.  Hence,  making  A"  the 
origin  of  moments : 

In  upper  chord  3  ;  J/  =  75,000  x  60  —  2  x  30,000  x  30=2,700,000 

ft.  lbs. 

These  operations  verify  a  previous  observation  to  the 
efTect  that  with  concentrated  loads  there  may  be  several 
maxima. 

Eq.  (10)  shows  that  the  greatest  moving  load  shear  in 
braces  i  and  2  is  42.2  tons  ;  hence,  multiplying  that  result  by 
its  tangent,  0.333,  ^"d  dividing  the  preceding  greatest  mo- 
ments by  the  depth  of  truss,  i.e.,  20  feet,  the  following  mov* 
ing  load  chord  stresses  are  found : 


RAIL  IVA  V  BRIDGES. 


49 


In  upper  did  i  ; '- — ^'^"^  =    84,271  lbs.=  42.14  tons. 


20 


(<     <(         (< 


«     <(         << 


^._  2,568,335^  128,417   "    =64.21     " 
=  135,000   "    =  67.50    " 


3;- 


20 
2,700,000 


20 


In  lower  cJi  d  \  \  —      42.2        x      0.333        =14.06    " 
1,936,400 


"    2;- 


20 


96,820  lbs.  =  48.41     " 


<<        <<  a 


3;-?^°=. 3^.9^4"    =66.46    " 


K'5). 


The  resultant  chord  stresses  are  found  by  adding  groups 
(4)  and  (5)  to  (15),  as  follows  : 

Upper  chord  (i)  =  —(14.79  +  42.14)  =  —  56.93  tons. 
"      (2)  =  -(23.58  +  64.21)  =  -87.79     " 
"      (3)  =  -(26.37+67.50)  =  -93.87     " 


(16). 


Lower  chord {\)  =  +(  5.24-1-14.06)  =  +  19.30  tons. 
"      (2)  =  +(18.03+48.41)  =  +66.44     " 
"      (3)  =   +(24.72  +  66.46)  =  +91.18     " 


Groups  (14)  and  (16),  therefore,  give  the  resultant  maximum 
stresses  in  all  members  of  the  truss. 

Those  web  members,  such  as  i,  3  and  5,  which  sustain  com- 
pression, are  called  "posts  "  or  "  struts,"  while  those,  such  as 
2,  4,  6  and  14,  which  sustain  tension,  are  called  "  ties." 

If  the  truss  be  divided  through  CD  and  either  LK  or  KH, 
it  is  seen  that  more  than  three  members  must  be  cut;  but  if 
that  number  is  exceeded,  it  is  known  from  the  first  principles 
of  statics  that  the  stresses  must  become  indeterminate. 
Hence,  when  coiinterbraces  are  introduced,  indetennination 
always  results.     If  provision  is  made  for  one  system  of  legiti- 


50  SPECIAL  NON-CONTINUOUS    TRUSSES. 

mate  stress   analysis,  however,  the  safety  of  the  structure  is 
assured. 

Only  one  point  more  needs  passing  attention  before  the  ex- 
amination of  the  next  case.  It  has  been  stated  in  the  course 
of  the  demonstrations  that  the  stress  in  certain  members  is 
tension,  and  compression  in  others.  In  web  member  4,  for 
example,  let  it  be  desired  to  determine  the  kind  of  stress.  It 
has  been  seen  that  when  the  greatest  main  web  stress  exists 
in  that  member,  the  reaction  at  R  is  32.058  tons,  and  it  is 
evident  that  it  is  directed  upward.  At  the  same  time  the 
live  load  resting  at  M  is  4.32  tons  and  is  directed  down.  The 
difference  of  these  forces  is  an  upward  ^&zx  oi  27.74  tons. 
Hence,  if  the  truss  is  divided  anywhere  between  BM  and  CL, 
this  shear  will  tend  to  move  the  left  portion  (between  the 
line  of  divisions  and  R)  upward  and  past  the  right  portion  ; 
/.  c,  it  will  tend  to  increase  the  distance  between  B  and  Z, 
and,  consequently,  produce  tension  in  web  member  4.  The 
general  principle  then  is  to  determine  the  effect  of  the  re- 
sultant external  forces  on  the  distance  between  the  extrem- 
ities, or  any  other  two  points  in  the  axis  of  the  member;  if 
the  tendency  is  to  increase  this  distance,  the  resulting  stress 
will  be  tension,  and  compression  if  the  reverse  is  the  case.  In 
trusses  with  parallel  chords,  after  a  very  little  experience,  the 
kind  of  stress  in  any  member  may  readily  be  discovered  at  a 
glance,  but  "in  many  structures  with  curved  or  polygonal  out- 
lines, resort  must  be  made  to  the  general  principle  stated 
above,  whicli  will  be  more  thoroughly  given  hereafter.  This 
simple  statement,  however,  is  all  that  is  needed  here. 

Art.    10.— Single    System    of  Vertical  and  Diagonal    Bracing. — Verticalc 

in    Tension. 

This  form  of  truss  when  built  with  timber  compression 
members,  has  long  been  known  as  the  Howe  truss;  The  skel- 
eton diagram  of  the  structure  to  be  considered  is  shown  in 
Plate  I,  Fig.  3.  The  moving  load  will  be  supposed  to  pass 
along  the  upper  chord  ;  hence  the  bridge  is  a  "  deck"  struc- 
ture. The  following  are  the  principal  dimensions  and  fixed 
load  data : 


RAIL  WA  Y  BRIDGES.  5  I 

Span  =  98  feet.  Panel  length  =  14  feet. 

Depth  =  20  feet.  Number  of  panels  =  7. 

Upper  chord  fixed  load  =  385  lbs.  per  lineal  foot  per  truss. 

Lower       "         "  "     =  215     "       "        "         "      " 

Upper       "         "  "  per  truss  panel  =   /[''  =  2.70  tons. 

Lower       '^         "  "     "        "         "      —   W  =  1.50     " 

tan  ABC  =  0.7  sec  ABC  =  1.22. 

The  moving  load  will  consist  of  the  two  consolidation  loco- 
motives used  in  the  preceding  Art.,  the  weights  of  which  are 
shown  in  Art.  yy  ;  and  this  load  will  be  taken  as  passing 
from  N  towards  M. 

The  web  stresses  will  first  be  determined,  and  the  first 
counterbrace  needed  comes  first  in  order.  As  the  number  of 
panels  is  seven,  ;/  in  Eq.  (7)  of  Art.  7,  is  equal  to  7.  Let  it 
be  required  to  ascertain  whether  the  compression  counterbrace 
LK  is  necessary.  If  the  train  is  so  placed  that  U\  rests  at  L, 
the  bridge  will  carry  the  weights  {Wy  ■■•  W-)  =  70.5  tons,  or, 
35.25  tons  on  each  truss.  Now  7  Jl\  =  52.5  tons,  and  7  {IV^ 
-\-  W.-,)  =  136.5  tons.  Since  the  total  load  on  the  bridge  is 
found  to  lie  between  these  values,  by  the  principles  of  Art.  7 
it  is  placed  to  give  the  greatest  compressive  shear  in  KL  or 
tensile  shear  in  HP.  In  order  to  find  the  shear  by  Eq.  (8)  of 
Art.  7,  the  following  values  result  from  the  position  of  the 
moving  load  just  taken;  x  =  1.3  ft.,  JV^  —  JV-  and  W„_iz= 
]i\.  Since  /  =  98  and/  =  14,  Eq.  (8)  of  Art.  7  gives  by  the 
aid  of  the  tabulation  on  page  41  ; 

_         I  /2,4i4,9io         141,000  \        I        121,200 

98\        2  2  ^/       14  2 

=  8920  lbs.  =  4.46  tons. 

The  fixed  load  shear  in  the  same  panel  is  W  -\-  JV  =  4.2 
tons.  As  the  latter  is  less  than,  and  opposite  in  kind  to  that 
of  the  moving  load,  the  counter  post  or  strut  KL  must  be 
introduced.  Since  the  difference  in  these  shears  is  very  small, 
it  is  evident  that  no  counterbrace  between  KL  and  O  is  needed, 
and  that  conclusion  may  easily  be  verified. 


52 


SPECIAL    NON-CONTINUOUS    TRUSSES. 


By  proceeding  in  precisely  the  same  manner  for  the  shears^ 
in  the  other  panels,  the  following  quantities  are  found  for  in- 
sertion in  Eq.  (8)  of  Art.  7,  when  Eq.  (7)  of  that  Article  is 
satisfied ; 


For  greatest  shear  in 
KL  .  .  .  W^  at  L  . 
HG...  W,  ''  N. 

F£. . .  in  "  ^ . 

DC  ..  .  JV^  "  D  . 

BA  .  ..  Ws  "  B  . 


lVn=W,    . 

.  .x  =  i.3. 

.  w,._,  =  iv,. 

IV,  =  w,  . 

.  -r  =  4.8  .  . 

.  w,._,^w,. 

IVn  =  Wn  . 

.  X  =  \.S  .  . 

.  w,,._,  =  iv,. 

W,^  =:  W,,   . 

.  .X  =6.S. 

.  iv,._,  =  w,. 

IV,  =  w,, . 

.  X  =  2.$  . 

.  w,._,  =  w,. 

That  Eq.  (8)  then  gives 


Shear  in  KL  = 


2,414,910       141,000 


X   1.3 


I        121,200 
X  ~    4.46  tons. 


14 


..       >.^g^ir3>967.86o^i7i^         gH^^   121 

981  2  2  J        14  ; 


200 


EE=  - 


•'  DC  = 


BA- 


7,009,740      210,000 


2  2 

10,431,240      282,000 


X    1.8    I : 

14 


X  6.8 


I 

—  X 
14 


2 

121,200 
2 

345.450 


10.05 


16.68 


-  25.33 


"15,631,350        327,000 

•^    -^^    +  —^ X  2.5 


I  345,450 

-  X  — — —=  35-79 
14  2 


The  shears  in  the  inclined  web  members  only  have  been 
given,  because  each  pair  of  braces  that  intersect  in  the  chord 
not  traversed  by  the  moving  load  take  their  greatest  stresses 
with  the  same  position  of  moving  load.  Braces  2  and  3,  4 
and  5,  etc.,  thus  go  together  in  pairs. 

The  resultant  web  stresse.s,  by  the  aid  of  the  preceding 
results,  will  then  take  the  following  values: 


Brace  i 
"  3 
"  5 
"      7 


-[3(^F+  W^)  +  35.79]  X  1.22  =  —  59.04  tons. 

•[2(W^  4-  M^^)  +  25.33]  X  1.22  =— 41.15     " 

-[    W+  W    +  16.68]  X  1.22=  -25.48     '' 

—  10.05  X  1.22  =  —  12.26     " 

-  4.46  X  1.22  =   -     5.44      " 


RAILWAY  BRIDGES.  53 

Brace  2  ..  .  2^F^  +  3^F+25.33  =  +  35.23  tons. 
"      4  .  .  .     W^  ^  2IV+  16.68  =^  +  22.38     " 
''      6  .  .  .  jr+  10.05  =  +  11.55     " 

The  positions  of  the  moving  load  for  the  greatest  chord 
stresses  are  found  by  the  aid  of  Eq.  (14)  of  Art.  7,  and  result 
in  the  following  quantities  : 

For  upper  chord 


3  • 

.  W-  2iiF  .  . 

•  w;  =  w,, . 

.  x=  2h.  . 

. .  ^.-1  =  ^K 

2  .  . 

.  w,  "  n .. 

.  w^  -  w,, . 

.  X  =  4  "  . 

■ .  w,._,  =  VV, 

I  . 

.  W,  "  B 

The  greatest  stress  in  upper  chord  i  occurs  with  the  maxi- 
mum shear  in  BA,  and  is  found  by  taking  the  product  of  that 
shear  by  the  tangent  of  its  inclination  to  a  vertical  line.  The 
shear  has  already  been  determined  to  be  35.79  tons,  and  the 
tangent  is  0.7.      Hence  ; 

Stress  in  upper  chord  i  =  —  35.79  x  0.7  =  —  25.05  tons. 

Eq.  (15)  of  Art.  7  then  gives  by  the  aid  of  the  tabulation  on 
page  41,  the  following  bending  moments: 


3 

Upper  chord  3  =_ 


2 

2  =- 

7 


"13.862,310   312,000 
2  2 


"13,862,310   312.000 


2,414,010 

'^  ^'^   ^1,896.754  ft.lbs. 


2 
1,020,450 


1,648,391  "  " 


As  the  depth  of  the  truss  is  20  feet,  the  moving  load  chord 
stresses  become : 

Upper  (3)  =  —  1,896,754  -^  20  =:  —  94,837  lbs.  =  —  47.42  tons. 
"       (2)  =  —  1,648,391  -=- 20  =— 82,420  ''     =—41.21     " 
"       (i)=  =-25.05     " 

By  combining  these  results  with  those  due   to   the  fixed 
load,  the  following  resultant  chord  stresses  are  found  : 

Upper  (i)  =  —  [25.05  +  3(1^+  W^)  X  0.7]  =  —  33.87  tons. 
"      (2)  =  -  [41.21  ■\-  S{W  ■\-  W)  X  0.7]  =  -  55.91     " 
"      (3)  =  -  [47.42  +  6{W  +  W)  X  0.7]  =  -  65.06     " 


54  SPECIAL   NOA'-COA'TIiVrOUS    TRUSSED: 

It  is  to  be  observed  that  the  upper  and  lower  chord  stresses 
are  the  same  in  pairs,  i.  c,  in  the  same  obhque  panel.  The 
reason  is  obvious.  If  a  panel,  as  DFEC,  be  divided  by  any 
line  cutting  DF,  DE  and  E.C,  it  will  be  evident  that  no  hor- 
izontal forces  whatever  exist  except  the  stresses  in  DF  and 
EC \  hence,  by  the  first  principles  of  statics,  those  stresses 
must  be  equal  in  amount  and  opposite  in  kind.  The  same 
result  holds  for  any  oblique  panel,  and,  indeed,  in  all  vertical 
and  diagonal  bracing  with  horizontal  chords  and  vertical  load- 
ing. The  stresses  in  one  chord  can  then  always  be  written 
from  those  in  the  other,  as  is  done  here  : 

Lower  (i)  =  +  33.87  tons. 
"       (2)  =--  +  55.91      " 
''      (3)  =  +  65.06     " 

In  reality  the  moments  remain  the  same  whether  the  upper 
or  the  lower  extremity  of  any  vertical  brace  is  taken  for  the 
moment  origin  ;  hence  the  equality  of  upper  aud  lower  chord 
stresses  in  the  same  oblique  panel. 

If  the  truss  becomes  a  through  one  (/.  r.,  with  the  load  on 
the  lov/er  chord),  the  same  pairs  of  web  members  do  not  inter- 
sect in  the  unloaded  chord,  hence  a  different  position  of  the 
moving  load  must  be  taken  for  the  greatest  shears  in  half  the 
braces.  A  simple  inspection  of  the  diagram  will  show  at 
once  that  the  position  of  the  moving  load  for  the  greatest 
shears  in  the  oblique  or  compression  braces  must  be  the  same 
whether  that  load  traverses  the  upper  or  lower  chord.  For 
the  vertical  braces,  however,  the  moving  load  must  be  ad- 
vanced in  the  lower  chord  at  least  one  panel  (more  in  some 
cases)  beyond  its  position  on  the  upper.  Hence  the  vertical- 
brace  stresses  will  be  greater  in  a  through  truss  than  those 
found  with  the  moving  load  on  the  upper  chord,  while  the 
stresses  in  the  oblique  braces  remain  the  same.  Conse- 
quently this  type  of  truss  is  better  adapted  to  the  deck  than 
the  through  form. 

Art.  11. — Single  System  of  Vertical  and  Diagonal  Bracing. — Verticals  in 

Compression. 

This  type   of  truss  is  very  common  in  American   bridge 


RAILWAY   BRIDGES. 


!); 


practice.  Its  compression  members  are  the  shortest  possible, 
and  its  details  are  simple  in  character,  and  both  those  fea- 
tures are  conducive  to  economy.  The  skeleton  diagram  ta 
■which  reference  is  to  be  made  is  given  by  Fig.  i,  of  Plate  II. 
The  bridge  is  supposed  to  be  a  "through"  structure,  hence 
the  floor  system  will  rest  on  the  lower  chord.  The  loads  and 
stresses  in  this  Art.  will  be  given  in  pounds.  The  principal 
dimensions  and  fixed  load  data  are  as  follows : 

Span     =  184'  iig"  Panel  length  =  20' 6g" 

Depth  =    27     o"  Number  of  panels  =     9 

Upper  chord  fixed  load  =  230  lbs.  per  lin.  ft.  per  truss. 
Lower  "         "         —  533        "  "  " 

Upper  "         "         per  truss  panel  —  IV  —    4726  lbs.. 

Lower  "         "  ''        "  =IV  =  10953   " 

1 5679  lbs. 

ta?i  ABL  —  0.761  sec  ABL  ~  1.26 

The  moving  load  will  consist  of  a  train  of  two  coupled  con- 
solidation locomotives,  each  with  concentrated  weignts  as 
shown  in  the  diagram  below,  followed  by  a  uniform  train  of 
2,240  pounds  per  lineal  foot.  It  will  be  taken  to  move  from 
right  to  left. 

The  moving  load  is  comparatively  light,  hence  the  fixed 
load  is  also  assumed  rather  low. 

As  usual,  Eq.  (7)  of  Art.    7  will  be   used   in   determining 


4^ 


2:M0l>s.  pt-rLiii.  ft. 


the  positions  of  moving  loads  for  the  greatest  shears,  and, 
hence,  for  the  greatest  stresses.  For  a  reason  that  will  here- 
after appear,  the  greatest  shear  for  every  panel  in  the  truss 
will  be  found.     The  application  of   Eq.  (7),  Art.  7,  will  giv*' 


56 


SPECIAL   NON-CONTINUOUS    TRUSSES. 


the    following   quantities   for   use    in    Eq.   (8)    of   the  same 
Art.: 


Max.  sJiear 
OP  . 
MN. 
FG  . 
EH . 
DI  . 
CJ  . 
BK  . 
AB  . 


in 


W^at  P  .  . 

.  IV„=W,      ...       ^=4.1 

W.  "  i\^  .  . 

''    =IV,      .      .     .        x=     2.2 

"    "  6^  .  . 

•    "   =  ^12     •     •     •        ;r  =     I  .o 

"    ''  H  .  . 

"    =  J'Fig      .      .      .          ;r  =     2.0 

w,  -  /  .  .  . 

*'  =IV]                    2x=    9.0 

U          U        CY 

''   —W     Uniform  2x  =  29.5 

"    "  A'  .  . 

"   =  W  '       load,     2^  =:  50.2 

"    "  z  .  . 

"  =  IV  ]                    2x  =  70.75 

ft. 


,  The  tabulation  to  be  used  in  connection  with  Eq.  (8)  and 
(15)  of  Art.  7,  is  given  below  and  is  formed  precisely  like 
that  in  Art.  9. 


I 

12,000  X 

7.666 

92,000 

2 

32,000  X 

4 

5S3 

146,666 

238,666 

3 

52,000  X 

4 

25 

221, coo 

459,666 

4 

72,000  X 

4 

583 

329.976 

789,642 

5 

92,000   X 

10 

666 

981,332 

1.770,974 

6 

106,000   X 

4 

666 

459.333 

2,230,307 

7 

120,000  X 

5 

583 

670,000 

2,900,307 

8 

134,000  X 

4 

666 

625,333 

3.525.640 

9 

148,000  X 

9 

333 

1,381,333 

4,906,973 

10 

160,000  X 

7 

666 

1,226,666 

6,133.639 

II 

180,000  X 

4 

583 

825,000 

6,958,639 

12 

200,000  X 

4 

25 

850,000 

7,808,639 

13 

220,000  X 

4 

5S3 

1,008.333 

8,816,972 

14 

240,000  X 

10 

666 

2,560,000 

11,376,972 

15 

254,000  X 

4 

666 

1.185,333 

12,562,305 

16 

268,000  X 

5 

583 

1.496.333 

14,058,638 

17 

282,000  X 

4 

666 

1,316,000 

15,374.638 

18 

296,000  X 

3 

25 

962,000 

16,336,638 

Eq.  (8)  of  Art.  7  gives  by  the  introduction  of  the  quanti. 
ties  found  above,  and  by  the  aid  of  the  tabulation : 


Shear  in  OP  = 


185 


459,666         72,000 
-^^-^ 4--^-! X    4.1 


=      2,040  lbs. 


3,525,640      148,000 

J.J    J.    ^    +-±^ X     2.2 


"  MN=-— 

185 


FG   ^M     6,958.639^200,000^    ^^ 
185 '|_  2  2 


92,000 
2  X  20.55 


=       8,171 


17,110 


RAIL  WA  Y  BRIDGES. 


57 


Shear  in  EH  - -r- 

185 


DI 


"   CJ 


BK 


AB 


185 
I 

I 

I 


11,376, 

2 

972 

254,000 

+    ^^           X     2.0 
2 

9^°^  =  29.884  '• 

2x20.55 

16,336,638 
2 

612,160 
+         2          ""     '^•^ 

•^3«.^«'=   45.783" 

2X20.55 

1 1 

658,080 
+    %         X  14.75 

-                 "               =      63,735      " 

" 

2 

—                  "              =      86,132      " 

" 

750,480      - 

+  ^     ;         X  35.38 

-                 "              =110,105      " 

Since  the  fixed  load  shear  in  HM  is  15,679  lbs.  it  is  seen 
that  counterbrace  10  is  the  first  counter  needed.  Combining 
the  fixed  load  shears  with  those  above  due  to  the  moving 
load,  the  following  resultant  stresses  in  the  inclined  web  mem- 
bers are  found  : 


.Stress  (10)  =  +  (   17,110 

"         (9)  =  +  (  29,884  )  X 

(7)=  +(  45,788  +  PF+  W)  X 
(5)  -  +  (  63,735  +  2W+2W')  X 
(3)=  +(  86,i32  +  3?'r+3J^0  X 
(i)z=: -(110,105  +  4?F+4[r)  X 


)  X  1.26  =  +    2\, ^60  lbs. 

=  +    37.654  " 

=  +    77,445  '' 

=  +  119,817  " 

=  +  167,800  " 

=  -  217,750  " 


Eq.  (21)  of  Art.  7  shows  that  if  [^4  be  placed  at  the  feet  of 
the  vertical  brace  2,  that  member  will  take  its  maximum 
moving  load  stress,  which,  by  Eq.  (19)  of  the  same  Art.  takes 
the  value : 


(2) 


4.05    X   6,000  4-  (11.72  +  16.32  4-  15.97)   10,000  +    2.92    X    14,000 
20.55 

+  10,000  =  +  34,600  lbs. 


The  stresses  in  the  vertical  braces  become  : 

Stress  (8)  =  -  (29,884  +     H'*)  =  -  34,61 1  lbs. 

''       (6)  =  -(45,788  +     f r +  2  fF)  =-66,193    " 
"      (4)  =  - (63,735  +  2?f^ 4- 3^)  =-99,819   " 

"       (2)  =  + (34,600  +         +     ^)=  + 45,553    " 

These  complete  the  greatest  stresses  in  the  braces. 


58 


SPECIAL   NON-CONTINUOUS    TRUSSES. 


Eq.  (14)  of  Art.  7  gives  the  following  values  for  the  posi- 
tions of  moving  load  for  the  greatest  stresses  in  the  chords: 

Upper  land  \  .  .  .  W^„_i=  IV^  ■  .  .  lVn  =  72,Soo  .  .  .  2x  =65.  //f. 

2 ..."     =  iVg  ..."  =82,300 ..."  =73.5  " 

I  .  .  .     "      =  H^5  .  .  .  "  =78,000  ..."    =69.7  " 

The  "2.f  "  shows  the  distance  covered  by  the  uniform  load. 
Eq.  (15)  of  Art.  7  then  gives  the  following  maximum  bend- 
ing moments  by  the  aid  of  the  tabulation  already  given : 


For  upper  3  and  4 


J/=^ 


16,336,638      737>6oo 

-  \  -        X  32.5 


9 
6,958,639 


5,478,153//.  lbs. 


M  =  - 


16,336,638  ,  756,640 


+ 


4,906,973 


36.8] 


=  4,910,013//.  lbs. 


M=- 


16,336,638  ^  748,130  ^ 


1, 770,97  A 


=  3,830,813//.  lbs. 


The  moving  load  stresses  in  lower  chord  i  and  2  are  found 
by  taking  the  product  of  the  greatest  shear  in  brace  i  by  its. 
vertical  tangent : 

Lower  (i)  and {2)  —  1 10,105  x  0.761  =  83,790  lbs. 

Since  the  depth  is  27  ft.  the  greatest  moving  load  upper 
chord  stresses  are  the  following: 

Upper  (3)  and  (4)  =-  5,478,153  -=-  27  =  -  202,900  lbs. 
(2)  =-4,910,013  H-  27  =-  181,850    " 
(i)  =-  3,830,813  -^  27  =-  141,900    " 

Since  [W +W')  tan  ABL  =  11,932,  the  resultant  upper 
chord  stresses  become : 


RAIL  WA  Y  BRIDGES.  59 

Upper  (i)  =— (141,900  +     7  X  11,932),=  -  225,424/^5. 

(2)  ==— (181,850  +    9x  11,932;  =— 289,238    " 
"  (3)^««'(4)  =—  (202,900  +  10  X  11,932)  =—  322,220   " 

In  the  lower  chord,  the  resultant  stresses  are  : 

Z,<?w^r  (i)  <a:«^(2)  =  +  83,790  +  4  X  11,932=-)-  131,520 /^J. 

(3)  =  =+  225,424    " 

"                      (4)  =  =  +  289,238    " 

"                    (5)  =  =+  322,220   " 

These  values  complete  the  resultant  stresses  in  the  various 
members  of  the  trusses,  and  they  will  be  used  hereafter  in 
making  the  complete  design  of  this  bridge. 

If  the  moving  load  traverses  the  upper  chords  of  the 
trusses,  the  same  pairs  of  braces  as  before  will  not  take  their 
greatest  shears  together.  The  stresses  in  their  inclined  braces 
will  not  in  any  way  be  changed,  but  it  will  be  necessary  to 
advance  the  moving  load  by  at  least  one  panel  beyond  the 
positions  taken  in  the  through  bridge,  in  order  to  determine 
the  greatest  shears  in  the  vertical  braces  of  the  deck  truss. 
Hence,  by  changing  the  bridge  from  the  "through"  type  to 
the  "  deck,"  the  vertical  braces  will  carry  considerably  in- 
creased stresses,  and  as  they  are  in  compression  the  weight 
of  the  bridge  will  be  materially  increased.  Hence,  this  truss 
is  best  adapted  to  carrying  the  moving  load  along  its  lower 
chord. 

Art.  12. — Two  Systems  of  Vertical  and  Diagonal  Bracing. — Verticals  in 

Compression. 

The  style  of  truss  shown  in  Fig.  2,  PI.  II.,  next  to  be 
treated,  was  at  one  time  more  common  than  any  other  in 
American  bridge  practice.  With  increased  facilities  for  fab- 
ricating and  handling  large  bridge  members,  it  has  been  pos- 
sible to  extend  the  use  of  single  systems  of  triangulation  to 
much  longer  spans  than  formerly.  In  this  manner  the  am- 
biguity of  the  double  system  is  avoided,  and  thus  analytical 
excellence  is  combined  with  advantages  of  production.     In 


6o  SPECIAL   NON-CONTINUOUS    TRUSSES. 

long  spans,  however,  the  double  system  is  still  frequently 
used,  and  if  properly  designed  it  is  not  so  objectionable  as 
might  at  first  seem  to  appear. 

It  should  always  be  arranged  with  an  even  number  of 
panels,  as  the  stress  ambiguity  is  then  reduced  to  an  unim- 
portant matter.  In  order  to  show  the  extent  to  which  am- 
biguity may  arise,  an  odd  number  of  panels  has  been  selected 
in  the  present  example.  As  has  already  been  shown,  the 
method  of  Art.  7  cannot  be  applied  to  a  double  system  of 
triangulation.  It  is  only  possible  to  assume  that  each  sys- 
tem acts  as  an  independent  truss,  and  to  determine  by  trial 
the  greatest  possible  concentrations  at  the  head  of  the  train, 
and  in  one  system,  and  consider  such  concentrations  the  head 
of  the  moving  load  in  that  system. 

There  will  be  two  equal  concentrations  represented  byi^^'at 
the  head  of  the  moving  load  in  each  system,  while  those  that 
follow  will  uniformly  equal  w.  Ordinarily  no  two  concentra- 
tions will  be  exactly  equal,  but  the  assumption  is  sufificiently 
accurate. 

The  truss  under  consideration  is  composed  of  two  systems 
of  right-angled  triangulation,  shown  in  Figs.  3  and  4  of  PI.  II. 

Before  passing  to  the  computations  it  is  well  to  observe 
that  although  the  action  of  the  loads  in  one  system  may  be 
considered  as  taking  place  independently  of  the  actions  of 
the  loads  in  the  other;  at  the  same  time  equal  loads  sym- 
metrically placed  in  reference  to  the  centre,  though  resting  on 
different  systems  of  triangulation,  may  be  considered  counter- 
balanced. The  web  stresses  due  to  the  fixed  load  will  be  de- 
termined on  the  supposition  that  the  web  members  shown  by 
the  dotted  lines  do  not  exist. 

The  data  to  be  used  are  given  below : 
Span  =210  feet.     Depth  of  truss  =  26  feet. 

Number  of  panels  =    15  Panel  length     =  14     " 

W  (upper)  =    9100  lbs.  =  4.55  tons  =    650  lbs.  per  foot. 
W^  (lower)  =  14000   "    =  7.00     *'     =  1000    "      "       " 
2^;  =  13  tons.  w'  =  20  tons. 

e  ^w'  —  tv  ■=  y  tons. 


TIVO   SYSTEMS  OF  BKACING.  6l 

Angle  QNO  =  a  Angle  MNO  =  /3 

tan  a  =  1.077  sec  a  —  1.47 

tan  (3  =  0.538  sec  ^       .         =1.136 

The  excess  e  will  be  taken  at  four  panel  points  as  before  ;  it 
will  also  be  used  in  determining  the  chord  stresses. 

The  counterbrace  16  is  the  first  one  needed.  Carrying  the 
moving  load  on  the  bridge  from  R,  panel  by  panel,  the  great- 
est web  stresses  are  found  to  be  the  following : 

In  brace  16.  .^  za  sec  ck  +  \^  e  sec  a  =  +    22.15  tons. 

"       "      IS.  .\izi^  sec  a  +  \j;  e  sec  a  =  +    28.62      " 

•"       "     \d,..\\%v  +^e+lV  =-    19.62      " 

"       "        I .  . (f I  zu  +  ^e)  sec  a  +  {IV+  W)  sec  a 

=  +     52.06      " 


i<        << 


iT,..\^zu  +  \^e  +  W                           =  -  24.02  " 

2 .  .  (fl  zv  +  \^e  +  W  +  W)  sec  a   =  +  59.80  " 

3- ■H'-^  -^  H^  +  2W+  W              =-  39.97  " 

4.  .  !f^  w  +  \^e  +  2  {W+  IV')\  sec  a 

=  +  84.53  " 

S..\\w  +  \^e  +  2W  +  W              =—  45.23  " 

6..]ff  w  +  f|r  +  2{W+  W')\  sec  a 

=  +  93-54  " 

7.  .^zv  +  ^e  +  slV+  2W'            =-  62.05  " 

8. .  Iff  zv  +  ^e  +  i{W+  IV')}  sec  a 

=  +  119-53  " 

g..^iu  +  ^e  +  slV+  2W'           =-  68.18  " 

io..mzv  +  f*e  +  s  {W+  W')\  sec  ^ 

=  +  100.33  " 


52  SPECIAL    NOA'-CONTrNUOUS   TRUSSES. 

In  brace  w  .  .w'  +  W  =  +     27.00  tons. 

"      12.  ./{w  +  W+  W)  sec  ^  +  {^e sec /i 

=  —  221.73      " 

The  stresses  in  each  system  of  triangulation  are  found  by 
virtually  taking  that  system  as  a  single  truss  supporting  only 
the  weights  at  the  apices  belonging  to  it. 

The  greatest  chord  stresses  will  be  obtained  by  supposing 
the  train  to  cover  the  entire  bridge,  with  the  four  excesses  e 
at  panel  points  1,  2,  3,  and  4. 

Greatest  stress  in  a  =  t,  {w  +  W  +  W)  [2  tan  fi  +  {tan  fi 
+  tan  a)\  ^-  {w  +  W  +  W)  tan  /i  +  c  {4  tan  fi  +  tan  a)  + 
\e  2  tan  ti  —  —  236.51  tons. 

Here  it  should  be  explained  that  since  \%c  —  'i^\e  is  found 
in  the  reaction  at  R,  the  three  r's  at  panel  points  i,  2,  and  3, 
and  \  of  that  at  4  may  be  taken  as  passing  directly  to  A', 
while  I  of  the  ^  at  4  passes  to  M  through  5',  2',  16,  14,  i,  3, 
etc.     Counterbrace  16  thus  comes  into  action. 

Greatest  stress  in  <^  =  [2  (w  +  W  -\-  W)  +  ^  ^]  tan  a  + 

236.514  =  —  291.91  tons. 

"  "       "  <:  =  2  (w  +  W^  +  W)  tan  ix  + 

291.907  =  -  344.79     " 

"  "        ''  d=\{w  +  W ^  W)  -\e\  tan  a  + 

344.787  =  -  366.20     " 

"  "        ''  e  =  {w  +  W  \  W)  tan  a  + 

366.201  =  —  392.64     " 

The  panel  stresses  in  r,  /",  and  g  will  be  the  same ;  and  if 
the  loading  were  uniform  over  the  whole  bridge,  the  panels  e, 
f,  g,  h,  and  k  would  all  be  subjected  to  the  same  stress. 

Greatest  stress  in  /  and  m  =  [y  {w  +  W  -\-  W)  +  3^  r]  tan  fi 

—  +  105.01  tons. 

«  "        "  «  =  [3  (w  +  W^  +  W)  +  \\e'\tan  ft  ^ 

105.009  :=    +    149-65        " 


TWO   SYSTEMS  OF  BRACING.  63 

Greatest  stress  in  ^  =  [3  (w  4-  W  +  W)  +  c]  tan  oc  4- 

149.654  —  +  236.51  tons. 

"  "        ''  p-  [2  {w  +  W  +   W)  +  \  r]  tan  n-  + 

236.513  =  +  291.91      " 

"  "        "  ^  =  2  (w  +  IV  +  W)  tan  ct  + 

291.906  =  +  34479     " 

"  "        "  r  =  {w  f   W^  +  W  -  *  r)  tan  a  + 

344.786==  +  361.17     " 


<<  (( 


"  s  =  {w  +  IV  +  JV  }  tan  n  + 

361.174=  +  387.61     " 

It  is  to  be  noticed  that  diagonally  opposite  panels  in  the 
upper  and  lower  chords,  up  to  the  counterbrace  16,  beginning 
with  the  panels  a  and  0,  are  subjected  to  the  same  amounts 
of  stress,  but  of  opposite  kinds. 

If  the  loading  were  uniform  over  the  whole  bridge,  this 
equality  of  the  pairs  would  continue  to  the  centre  ;  also,  the 
stresses  in  the  panels^, /,^,  //,  k,  and  s  would  be  equal  to 
each  other. 

If  the  end  posts  were  vertical,  there  would  be  obvious 
changes  in  the  stresses  of  the  panels  /,  w,  and  n  (that  in  / 
would  be  zero).  The  upper  chord  panel  stresses  would  not 
be  changed. 

The  whole  truss  in  Fig.  2,  PI.  II.,  is  composed  of  the  two 
systems  of  triangulation  shown  in  Figs.  3  and  4,  and  each  of 
these  is  to  be  considered  separately  in  checking  the  chord 
-Stresses  by  the  method  of  moments.  Denote  by  (R)  and 
-(R")  the  reactions  at  the  points  indicated  by  the  same  letters 
in  Figs.  3  and  4,  then  divide  the  total  load  supported  by 
each  system  into  two  parts,  according  to  the  principle  of  the 
Jevcr,  and  there  will  result : 

{R')  =  If  [7(i£/  +  JV+  JV')]  +  i|  .  2^  =  103.7867  tons. 
{R")  =  ^4  [7  (w  -f  IV  +  IV')]  +  ff  •  2  ^  =    91.3966  tons. 


64  SPECIAL   NON-CONTINUOUS   TRUSSES. 

If  the  diagonals  are  in  tension,  according  to  this  value  of 
{R),  D'L'  should  be  drawn,  and  not  K'E'.  The  latter  is 
taken,  however,  for  a  reason  that  will  appear  presently. 

The  sum  of  {R)  and  {R')  is  just  equal  to  the  total  reaction 
at  R  in  Fig.  2,  as  it  ought  to  be. 

Indicate  by  {BC),  {DC'),  {d),  etc.,  the  stresses  in  the  panels 
represented  by  those  letters.  Taking  moments  about  H  and 
G  respectively,  there  result : 

{BC)  =  -{HK)  =  [{R)xRH-2{w'+VV^W){GH+\FG)']^d. 
{AB)  =  -{GH)=l{R)xRG-{w'+W+W')FG]^d. 

Also,  taking  moments  about  K'  and  H' : 

(C'D')^[{R')  X  R'K' -2{za'  +  W+  W)  (H'K'  +  ^G'H')]^d. 
{B'C')  =  -{HK')  =  [{R')'kR"H'-{iv'^  W+  W)  G'H')-=rd, 

Similar  expressions  will  give  the  chord  stress  in  every  panel 
of  Figs.  3  and  4;  and  having  found  these,  the  resultant 
stresses  in  Fig.  2  are  simply  the  sums  of  the  proper  pairs 
taken  from  Figs.  3  and  4. 

Thus,  {d)  =  (BC)  +  (CD) 

(e)    =  {CD)   +  {CD') 
(0)   =  (GH)  +  (G'ff') 
etc.  =    etc.    +     etc. 

This  system  of  determina.tion  by  moments  may  be  applied 
to  any  truss  with  parallel  chords,  however  many  systems  of 
triangulation  there  may  be. 

The  method  also  applies  to  any  irregular  loading,  for  the 
stresses  due  to  each  panel  load  may  be  found  separately,  and 
the  sum  caused  by  all  taken. 

Web  stresses  may  also  be  checked  by  the  same  method, 
since  the  increment  of  chord  stress  at  any  panel  point  is  equal 
to  the  sum  of  the  horizontal  components  of  the  stresses  in 
the  web  members  intersecting  at  the  panel  point  in  question. 
Such  a  check,  however,  is  a  very  tedious  one. 

Applying  the  above  equations  to  C ' D'  in  Fig.  4  : 

{C'D)  =  (91.397x84  -  2  X  31.55  X  42)  -4-  26  =  193.35  tons.. 


TfVO  SYSTEMS   OF  BRACING. 


65 


Also,  to  CD  in  Fig.  3 


{CD)  =  (103.787  X98  —  2x31.55x70  —  24.55  X  28)  ^  26  = 

194.95  tons. 

But  the  sum  of  these  two  is  388.3  tons,  whereas  {e)  =  ^g2.64 
tons.  This  discrepancy,  not  very  great,  is  easily  explained. 
The  loading  (zu  +  IV  +  W)  is  counterbalanced  in  Fig.  2,  but 
is  not  in  Figs.  3  and  4. 

In  Fig.  2  all  the  load  on  the  left  of  the  centre  of  the  span, 
except  f  r  at  4  or  //',  is  assumed  to  pass  directly  to  R  (or 
R'  and  R").  Hence  in  Figs.  3  and  4,  to  be  consistent  with 
Fig.  2,  there  should  be  taken  : 

{R')  =  4 (w  +   W  4-  W)  +  2e  =  1 12.2  tons. 
\r')  =  3  (zu  +  IV  +  IV)  +  ^e=    82.983  tons. 

Introducing  these  in  the  general  formula: 

(e)  =  {CD')  +  {CD)  =  [1 12.2  X  98  +  82.983  X  84  -  31.55  X 
84  -  31.55  X  140  -  24.55  X  28]  -H-  26  =  392.75  tons. 

This  result  agrees  sufficiently  well  with  that  obtained  by 
the  trigonometrical  method. 

With  the  last  value  of  {R),  K'E  will  be  in  tension. 

It  is  thus  seen  that  v/ith  an  uneven  number  of  panels  a 
little  ambiguity  exists  both  in  reference  to  the  greatest  chord 
stresses  and  the  greatest  web  stresses,  when  there  are  two 
systems  of  triangulation.  TJiis  ambiguity  always  exists,  what- 
ever the  number  of  systems,  if  the  compotient  systems  are  not 
symmetrical  i?i  reference  to  the  centre  line  of  the  span,  and  it 
always  disappears  if  they  are  symmetrical  in  reference  to  that 
line. 

With  an  even  number  of  panels  in  the  span  and  two  sys- 
tems of  triangulation  no  ambiguity  exists. 

These  observations  in  reference  to  ambiguity  apply  as  well 
to  isosceles  bracing  as  to  vertical  and  diagonal. 

In  the  example  taken  there  are  only  two  systems  of  triangu- 
lation, but  precisely  the  same  method  is  to  be  followed  what- 


■  ^^  SPECIAL   NON-CONTINUOUS    TRUSSES. 

ever  the  number ;  in  determining  the  web  stresses,  each  sys- 
tem is  supposed  to  carry  those  moving  weights  only  which 
rest  at  its  apices,  and  the  same  is  true  in  reference  to  chord 
stresses  for  unsymmetrical  loading,  uniform  loading  being 
supposed  counterbalanced  for  either  stresses. 

The  slight  changes  to  be  made  for  an  overhead  bridge,  or 
for  verticals  in  tension  and  diagonals  in  compression,  are  evi- 
dent from  what  has  already  been  given  in  preceding  articles. 

It  is  seen  that  any  two  web  members  intersecting  in  the  chord 
not  traversed  by  the  moving  load  receive  their  greatest  stresses 
at  the  same  time  ;  the  principle,  indeed,  is  a  general  one. 

When  built  in  iron,  this  truss  is  frequently  called  the  Lin- 
ville  truss. 

Art.  13. — Truss  with  Uniform  Diagonal  Bracing — Two  Systems  of  Tri- 

angialation. 

This  truss  is  shown  in  PI.  X.,  Fig.  6,  and,  although  taken 
here  as  an  ordinary  pin  connection  bridge,  precisely  the 
same  method  of  calculation  is  to  be  used  for  a  "  lattice"  truss 
with  riveted  connections. 

No  locomotive  excess  will  be  taken,  but  a  heavy  moving 
load  of  uniform  density  will  be  assumed.  The  following  are 
the  data: 

Span  —  182  feet.     Depth  =  23  feet. 

Panel  length  =     13    "         Number  of  panels  =  14. 

Fixed  load  : 

W  (upper)  =  450  pounds  per  foot  =  2.925  tons  per  panel. 
W  (lower)  =  800       "  "       "     =  5.2 

Moving  load : 

w  —  2800  pounds  per  foot  =  18.2  tons  per  panel. 
Angle  AaB  ~  a. 


tan  a 

=  0.565. 

sec  a          =     1 .  1 5  • 

W  sec  a 

=  3.364  tons. 

W  sec  a  —    5.98  tons. 

w 

—  sec  a 

14 

=  1.500     " 

Wtana    =     I.653    " 

W  tan  a 

=  2.94      " 

wtan  (X     =  10.28      " 

TRUSS    WITH    UNIFORM  DIAGONAL   BRACING. 


67 


The  vertical  members  aB  and  tS  are  for  tension  only. 

The  moving  load  will  be  taken  as  passing  from  A  to  T,  and 
its  head  will  be  supposed  to  rest  at  the  various  panel  points 
in  succession,  in  the  determination  of  the  web  stresses. 

The  notation  for  the  stresses  is  one  which  will  frequently 
be  used  hereafter.  The  stress  in  any  member  is  indicated  by 
inclosing  in  a  parenthesis  the  letters  which  belong  to  it  in 
the  figure. 

Head  of  moving  load  at  D. 
(dF)  —  \\  W  +  W —-^w\  sec  ex  —  -{-  2.6  X  sec  a. 

Hence  the  stress  in  dF  v^WX  always  be  tension. 


{Ee) 


Head  of  moving  load  at  E. 
w  —  W  —  \W\  sec  a  ^  —  1.79  X  sec  a. 


Hence  the  stress  in  Ee  will  always  be  compression. 
The  web  stresses  desired  are,  then,  the  following  : 


{Ff)   =  - 

(  ^IV  + 

W  —  Jj  w)  sec  a           =  + 

7.15  tons. 

{eG)    = 

(     JV  + 

^W-^^^w)     * 

'              =  —. 

1-34     ' 

{Gg)   =  - 

( 

hW-\lw)     ' 

=  + 

16.32     ' 

{fm  = 

(  ^w 

-t\w)     ' 

'              =  — 

10.51     ' 

{Hh)  = 

(  iw 

+  \iw)     ' 

=  + 

26.99     ' 

igK)  =  - 

( 

^W+\lw)     ' 

'              =  — 

19.68     ' 

(Kk)  = 

(     w  + 

hW^-^w)     ' 

=  + 

37.66     ' 

{liL)    =  - 

(  ^w  + 

W  +  i|  w)     ' 

<              =  _ 

30.35     * 

(LI)    = 

(i^W  + 

W+^w)     ' 

'              =  + 

49-83     ' 

(W)   =  - 

(     w  + 

i^w^+fi^O    * 

'              =  — 

41.02     * 

iOo)    = 

(    2W'  + 

i^W^^w)     ' 

*              =  + 

62.00     ' 

(IP)    =  - 

(i^W  + 

2W+  ^w)     ' 

'              =  — 

53.20     ' 

(Pp)    = 

(2iW'  + 

2IV  +  ^zv)     ' 

*              =  + 

75.68     ' 

(oQ)    =- 

{2W'  + 

2^W  +  \^W)       ' 

=  — 

65.37     ' 

(Qt)    = 

(  3^'  + 

2^W+\lw)       ' 

'              =  + 

89-35     ' 

(ps)  =  - 

(2hW'  + 

3fF+||w)     ' 

'              =  — 

79.04     ' 

{tT)   =  - 

^(W  + 

W  +       w)     ' 

'                       :=   — 

197.24     ' 

(fS)    = 

ihW  + 

l\V+\iw 

=    + 

90.68     ' 

68 


srECIAL   NON-COXTIMOrs   TKrSSES. 


With  the  moving  load  covering  the  whole  bridge,  the  fol- 
lowing chord  stresses  are  found  : 


{ah)  =  —  (9*  JF'  +  9  ^F  +  9i  2£/)    tan  a 

{be)  ={ad)     -  s{lV'  +  W+w)     - 

{cd)  ={bc)     -4(     " 

{de)  ={cd)    -3(     " 

{ef)  ={de)     -2(     - 

ifg)  ={ef)     -     (     - 

{AB)  =  6i(     '' 

{BC)  ={AB)  +{2^lV'  +  2,lV+2Uv)  " 

{CB)  =  {BC)  +  S{W'  +  W+w) 

{DE)={CD)  +4(     " 

{EF)  ={DE)  +  3(     - 

(i^^;)  =(^i^)  +  2(     " 

{GH)=  {EG)   +     (     " 


—  —  141.46  tons. 
=  -215.83  " 

=  -275-33  " 

=  -319.95  " 

=  -  349-/0  '' 

=  -  364-57  " 

=  +    96.68  " 

=  +  134.69  " 

=  +  209.06  " 

=  +  268.55  " 

=  +  313-17  '' 

=  +  342.92  " 

-  +  357-80  " 


The  following  operations  constitute  a  check  on  the  accuracy 
of  the  chord  stresses. 

The  horizontal  forces  exerted  at  the  joints^  and  //,  respec- 
tively, are : 

if£)     —  \Wta71  a  =  —  365.40  tons, 

and  {GH)  +  ^{W  +  w)  ta7i  a—-^  364.41     " 

The  horizontal  force  exerted  at  either  one  of  these  joints^ 
as  found  by  the  moment  method,  is: 


T  {W  ^  IV+  w)  X  0.25  X  182  _ 
23  ~ 


364.5  tons. 


The  agreement  is  close. 

It  is  to  be  observed  that  {E/)  is  the  greatest  tensile  stress 
in  /iL,  also;  and,  on  the  other  hand,  that  {/iL)  is  the  greatest 
compression  stress  E/.  Similar  observations  apply  to  the 
pairs  of  members  eG,  kK\  Gg,  Kg\  fH,  Hh. 

These,  consequently,  are  the  only  web  members  which  need 
to  be  counterbraced. 

Precisely  the  same  methods  of  calculation  apply,  whatever 


COMPOUND    TRIANGULAR    TRUSS. 


69 


may  be  the  number  of  systems  of  triangulation  or  the  char- 
acter of  the  load,  or  whether  the  truss  be  a  through  or  deck 
one. 

If  Fig.  6  represented  a  deck  truss,  however,  the  compres- 
sive web  stresses  would  be  increased  and  the  tensile  ones 
diminished,  while  the  chord  stresses  would  remain  the  same. 
Since  the  increase  of  compression  in  any  web  member  would 
numerically  exceed  the  decrease  of  tension  in  the  adjacent 
one,  the  truss  is  better  adapted  to  a  through  load  than  a 
deck  load. 

This  truss,  particularly  with  only  one  system  of  triangula- 
tion, is  frequently  called  the  "triangular"  truss. 

Art.  14. — Compound  Triangular  Truss. 

A  very  economical  style  of  truss,  in  point  of  quantity  of 
material,  is  that  shown  in  Fig.  i  of  PI.  III.  The  truss  is  of 
the  ordinary  isosceles  bracing,  and  formed  of  two  systems  of 
triangulation,  but  a  half  of  the  floor  system  and  moving  load 
is  carried  by  verticals  directly  to  the  intersections  Ey  F,  etc. 

Half  the  weight  of  the  trusses  is  supported  at  the  apices  of 
the  main  systems,  as  H  and  M,  in  the  upper  chord,  and  half 
at  the  apices,  as  P  and  R,  in  the  lower  chord.  The  truss 
chosen  is  a  deck  or  overhead  truss;  consequently  half  the 
floor  system  and  moving  load  will  be  supported  by  the  verti- 
cals in  compression.  The  weight  of  the  floor  system  will  be 
taken  at  300  pounds  per  foot,  and  the  moving  load  taken  will 
be  a  uniform  one  made  up  of  a  load  of  heavy  engines  weigh- 
ing 2700  pounds  per  foot.  In  such  a  case  there  is  no  ex- 
cess e. 

The  following  are  the  data  : 

Length  of  span  =  200  ft.        Depth  of  truss      =  27.75  ft. 

Upper-panel  length  =  12.5   "         tan  CDL  —  tan  a  =    0.9 
Lower-panel  length  =  25.0  "         sec  CDL   =  sec  «=     1.345 

W    (upper)  =25     X    500  +  \2h  X  300=    8.125  tons. 
W^i  (middle)  =  12^  x    300  —     1.875     " 

W    (lower)  =25     X    500  =    6.25       " 

'w='w'  —  \2h  X  2700  =  16.875     " 


70 


SPECIAL    NON-CONTINUOUS    TRUSSES. 


The  middle  loads  W-^  or  w  are  applied  to  the  trussing  as 
follows.  The  adjoining  figure  represents  a  portion  of  the 
truss  in  question  as  indicated  by  the  same  letters  HMPR  (see 

figure  in  plate).  Any  weight  rest- 
ing at  K  is  carried  down  to  the  in- 
tersection, or  two  apices  A  and  B, 
and  the  proper  portion  of  each 
load  is  hung  at  each  apex.  In  the 
truss  in  question,  A  C,  in  the  ad- 
joining figure,  will  represent  \^  of 
the  weight  at  K,  and  BD  -^  of  the 
same  weight.  The  moving  load  is  supposed  to  pass  on  the 
bridge  from  A.  By  examination  it  is  seen  that  o  and  j-  are 
the  first  members  which  need  counterbracing.  The  head  of 
the  train  must  be  at  the  panel  point  between  /  and  r  for 
greatest  moving-load  stress  in  s,  and  at  the  panel  point  be- 
tween /  and  ^  for  that  in  o,  and  at  corresponding  positions 
for  other  web  members. 

Fixed-load  stress  in     s.  .=  ^(W  +  W^)  sec  a  =  —     6.73  tons. 


Moving-load  stress  in  s. 


Fixed-load  stress  in     o. 


=  (i  +  3  +  8  +  5;-'.y^f  a- 

=  +     12.05     " 

=  \{W'  +  W^)s€ca^  -^       5.46    " 


Moving-load  stress  in  0. 


(i  +  4  +  3  +  5  +  12)  — sec  a 


Fixed-load  stress  in     p .  .  =  \  W  sec  oc 


^7-71    " 
5.46    " 


Moving-load  stress  in/. 


w 


(1+  3  +  8  +  5  +  y)~^sec  a 


+     17.02 


Fixed-load  stress  in     r.  .  =  i  IV  sec  a 


+    4.20       " 


COMPOUND    TRIANGULAR    TRUSS.  ^j 

iv 
Moving-load  stress  in  r.  .  =  (i  +  4  +  3  +  5  +  12  +  7)  —  ^^^  « 

=  —    22.70  tons. 

Greatest  stress  in  i . .  =  (||  a'  +  h  W)  sec  a 

=  -    33.82    " 

"  "        "  2  .  .  =  (|i  «'  +  ^IV'+  i  PVi)  sec  a 

=  +    34-53    " 

=  +     26.9       " 

"       "  4-.=  {n^^''+  ^^+  ^J'K)sec  ct 

=  -    4147    " 

"       "  5--=  (H«'  +  H^'+  I'V,)  +   IV)  sec  a 

=  —    59.64  tons. 

"       "  S..  =  mw+^{lV'+2lV,)  +  lV)seca 

=  —    68.72  tons. 

"       "  6-.=  {nu'+  W-^  ^^+ii'K)  sec  a 

=  +    49.87  tons. 


<(       (< 


7..=  (Uw  +  hlV+   JV+   W,)seca 

=  +     58.93  tons. 

9.  .  =  (H  w  +  I J^  +   W+  W,)  sec  a 

=  -    86.88  tons. 

12 .  .  =  (II  2v  ^  IW  +  IV  +  i JV,)  sec  a 

=  —    97.38  tons. 

ll..=  mzv  +  IW+   IVi  +   JV)  sec  a 

=  +    77.13  tons. 

I0..=  mw+  ^W'+lW,  +   W)seca 

=  +    87.59  tons. 

13- •=  (W-^+  |f^'+  1^1  +  2W)sec  a 

=  —118.37  tons- 


72 


SPECIAL   NON-CONTINUOUS    TRUSSES. 


'Greatest  stress  in        i6 . .  =  (l^^  zv  +  IW'+  2IV1  +  2  W)  sec  a 

=  —  130.30  tons. 

"  "       "         1^..=  {^w -V  ^W  +  2W'+ iW^)sec  a 

=  +  105.79  tons. 

"  "       "         i^..=  {^.i-w  +  lW+2W'+ 2W;)seca 

=  +  117.66  tons. 

"  "       "         17.  .  =  V^8_  ^e/  +  2W+  2W'+2  Wi 

=  —  100.00    " 

"  "       "        i8..=  lV+w  =-    25.00    " 

The  stress  in   17  added  to  the  vertical  component  of  the 

: stress  in  16  is  equal  to  {Siv  +  4W  +  4IV1  +.3.^  IV ')  the  weight 

■of  the  truss  and  its  load,  as  it  should.     This  constitutes  a 

check  in  the  work  if,  as  was   done,  each  web  stress  is  found 

by  adding  a  proper  increment  to  a  preceding  one. 

For  the  greatest  chord  stresses  the  load  will  cover  the  whole 
bridge. 

Stress  in  a  or  d.  .=  {^^zv  +  pV+  2W'  +  2  l\'\)  tan  a 

"  —   78.75  tons. 

"    c  ord..=  (6  zv  +  3  W  +  3  TFi  +  3  W)  tmi  a  + 

78.75  =  -213.75     " 

''        "   r  or /. .  =  2  (2  zv  +  W  +  W^  +  W)  tan  a  + 

213-75  =  -  30375    " 

"        "  gor  h..=  {2ZV  +  W  +  JFi+  W)  tan  a  + 

30375  =  -  34875    " 


11.  a 


II  a 


. .  =  {4U'  +  2IV+  2 IVi  +  I  W)  tan  a 

=  +     87.19    " 

.  .  =3  (2w  +  IV  +  JVi  +  W)  tan  a  + 

87.19=  +  222.19    " 

"  /  .  .  =2  (2w  +  W  +   Wi  +   W)  tan  n  + 

222.19  =  +  312.19     " 


COM r OCX D    TRIANGULAR   TRUSS. 


n 


Stress  in  k.  .  —  {2iv  -^  W+  IV^  +  W)  tan  a  + 

312.19  =  4-  357.19  tons. 

In  determining  these  values,  it  is  to  be  remembered  that 
the  increment  of  chord  stress  at  any  panel  point  is  equal  to 
the  sum  of  the  horizontal  components  of  the  stresses  in  the 
web  members  intersecting  at  that  point. 

The  results  for  g  or  //  or  k  may  be  easily  verified  by  the 
method  of  moments.  Let  /  be  the  span  in  feet,  and  a  the 
depth  of  a  flanged  beam,  in  feet  also  ;  then  if  w  is  the  load 
per  foot,  the  flange  stress  at  the  centre,  as  is  well  known,  will 

be  -5-1-.  To  apply  this  to  the  present  case,  {zu  +  IV -\-  W^  +  W) 
must  be  written  for  lu,  and  /  and  d  have  the  values  respect- 
ively of  200.00  and  27.75.     Hence  -^-7-  =  360.4  tons. 

ou 

Now  since  the  resultant  stress  at  either  of  the  centre  joints 
is  horizontal  in  direction  for  a  uniform  load  from  end  to  end 
of  the  truss,  the  value  corresponding  to  the  above  will  be 
found  by  adding  to  the  stress  in  //  the  horizontal  component 
of  the  stress  in  brace  i,  for  the  supposed  uniform  load  ;  or  by 
adding  to  that  in  panel  k  the  horizontal  component  of  that  in 
brace  3. 

Horizontal  component  in  i  =i- j  tana—  11.25  tons, 

and  348.75  +  11.25  —  360.00  tons. 

W 
Horizontal  component  in  3  = tan  a  =  2.8125  tons, 


and  357-19  +  2.81  =  360.00  tons. 

Both  of  the  above  results  are  remarkably  satisfactory  verifi- 
cations ;  they  would  have  agreed  exactly,  but  0.9  is  not  the 
•€xact  value  of  tan  a. 

If  the  bridge  were  a  through  one,  the  general  method  of 


74 


SPECIAL  NON-CONTINUOUS    TRUSSES. 


calculation  would  be  exactly  the  same ;  the  slight  changes  in 
the  details  of  the  operations  are  sufficiently  obvious  after 
what  has  been  said  before. 

As  a  through  truss  there  would  be  some  saving  of  material, 
for  the  secondary  verticals  i8  would  be  in  tension. 

A  much  greater  saving  might  be  effected  by  using  inclined 
end  posts,  in  which  case  a  short  beam  or  girder  would  take 
the  place  of  the  end  panels  LC,  and  braces  15  would  be  verti- 
cal and  run  up  to  Z,  while  braces  18,  14,  and  17  would  be 
omitted  altogether. 

Art.  15. — Methods  ot  Obtaining  Stresses — Stress  Sheets. 

In  the  preceding  cases  the  analytical  expressions  for  the 
stresses  have  been  written  in  such  a  manner  as  would  seem 
best  to  show  in  detail  the  principles  by  which  they  are 
traced. 

In  practice  every  engineer  has  a  method  best  fitted  for 
himself  by  either  habit  or  taste. 

The  "  strain  sheet,"  or  properly  "  stress  sheet,"  is  almost  in- 
variably made  as  shown  in  the  plates.  A  skeleton  of  the 
truss  is  drawn,  and  along  each  member  is  written  the  great- 
est stress  belonging  to  it. 

Art.    16. — Ambiguity  caused  by  Coimterbraces. 

It  is  important  to  notice,  from  what  has  preceded,  that  a 
little  ambiguity  always  exists,  both  in  web  and  chord  stresses, 
near  the  middle  of  the  truss,  when  cotinterbraccs  are  used  in- 
stead of  counter  bracing.  This  arises  from  the  fact  that  even 
with  a  single  system  of  triangulation  it  is  impossible  to  divide 
the  truss  by  cutting  less  than  four  members,  which  is  equiva- 
lent, as  a  question  of  equilibrium,  to  having  four  unknown 
quantities  and  only  three  equations  by  which  they  are  to  be 
determined. 

This  ambiguity,  however,  has  been  shown  to  be  not  of  a. 
dangerous  character. 


CHAPTER   III. 

NON-CONTINUOUS   TRUSSES   WITH    CHORDS   NOT   PARALLEL. 
Art.  17. — General  Methods. 

The  determination  of  stresses  in  trusses  with  chords  that 
are  not  parallel  can  usually  be  accomplished  more  con- 
veniently by  either  a  combination  of  the  method  of  moments 
with  the  graphical  method,  or  the  graphical  method  alone, 
than  in  any  other  manner. 

So  long  as  three  members,  at  most,  are  cut  by  any  surface 
whatever,  dividing  a  truss  into  two  parts,  the  problem  of  the 
determination  of  the  stresses  in  these  members  is  determin- 
ate ;  for  in  such  a  case  the  problem  is  really  one  of  the  equi- 
librium of  any  system  of  three  forces  parallel  to  a  given 
plane,  for  the  solution  of  which,  as  is  well  known,  there  are 
three  equations  of  condition. 

The  matter,  however,  requires  a   little  attention    here,  in 
order    that    the    particular    kind    of    stress   (tension  or   com- 
pression) developed  in 
any  bar  may  be  known 
from    the    stress    dia- 
gram. 

Let  Fig.  I  represent 
a  portion  of  any  truss 
divided  into  two  parts 
by  the  plane  (it  might 
be  any  other  surface) 
AB\  let  i%  G,  and  H 
be  the  points  of  inter, 
section  of  this  plane 
and  the  three  mem- 
bers CK,  CD,  and  ED ;  and  let  2P  be  the  resultant  of  all  the 
external  forces  acting  on  that  portion  of  the  truss  lying  on 

75 


Fig.  I. 


76  NOI\'-CONTINUOUS    TRUSSES  IX  GENERAL. 

the  left  o{  AB.    The  external  forces  are  known,  and  the  stresses 
c,  /',  and  /,  in  CK,  CD,  and  ED  respectively,  are  required. 

Now,  any  one  of  those  stresses  may  be  determined  by  the 
method  of  moments,  if  the  origin  of  moments  be  properly 

located.     If  the  origin  be  taken  at  the 
point  of  intersection  of  the  lines  of  ac- 
tion   of  any   tzvo   of  the   stresses,   the 
moments  of  those  stresses  will  be  aero. 
SP  \^/'  Hence,  as  a   general    principle,  in 

order  to   determine  any  one   of  those 
unknown  stresses  by  the  method  of  mo- 


c 

^  ments,  the  origin  of  moments  is  to  be 

Fig.  2,  »  <b         y 

taken  at  the  intersection  of  the  lines  of 
action  of  the  other  tzvo ;  the  moment  of  the  third  unknown 
stress  can  then  be  placed  equal  to  the  resultant  moment  of 
the  external  forces,  giving  one  equation  with  one  unknown 
quantity. 

Suppose  in  Fig.  i  that  the  stress  in  CK  is  to  be  found  by 
moments.  D  is  the  point  of  intersection  of  CD  and  ED,  and 
consequently  is  the  origin  of  moments.  The  lever  arm  of 
that  stress  is  of  course  the  normal  distance  from  D  to  CK. 
The  kind  of  stress  in  CK  can  always  be  determined  by  the 
known  direction  of  the  resultant  external  moment.  If  the 
effect  of  the  moment  is  to  shorten  the  piece  in  question,  the 
resulting  stress  will  be  compression,  while  tension  will  exist  if 
the  effect  is  to  lengthen  the  piece. 

If  any  one  of  the  three  stresses  is  known,  either  of  the 
others  may  be  found  by  moments,  by  taking  the  origin  of 
moments  anywhere  on  the  line  of  action  of  the  third  force. 
This  method  is  illustrated  on  page  255. 

In  Fig.  I  it  will  be  assumed  that  the  effect  of  the  moment 
of  the  external  forces  on  the  left  of  AB  is  a  tendency  to  turn 
that  portion  of  the  truss  in  the  direction  of  the  curved  arrow, 
and  consequently  to  shorten  CK,  thus  producing  compression 
in  that  member.  Let  c  represent  that  compression,  then,  so 
far  as  the  portion  of  the  truss  on  the  left  of  AB  is  concerned, 
it  is  equivalent  to  an  external  force  acting  from  F  toward  Cf 
as  shown  in  the  figure  by  the  arrow  c. 


GEXERAL  METHODS. 


77 


Now,  it  is  known  from  rational  mechanics  that  if  the  sides 
of  a  closed  polygon  represent  a  system  of  forces  in  equilib- 
rium, the  lines  representing  the  forces  must  be  laid  off  in  the 
same  direction  around  the  polygon,  in  order  that  the  true 
direction  of  those  forces  may  be  represented.  If,  for  instance, 
the  lines  of  the  polygon  in  Fig.  2  represent  a  system  of  forces 
in  equilibrium,  the  forces  must  act  in  the  direction  shown 
by  the  arrow-heads  :  all  in  the  same  direction  around  the 
polygon. 

The  portion  of  the  truss  on  the  left  of  AB,  Fig.  i,  is  held 
in  equilibrium  by  the  external  forces  acting  upon  it,  and  the 
three  stresses  acting  at  F,  G,  and  H.  The  amount  and  direc- 
tion of  the  stress  acting  at  F  have  already  been  found  ;  they 
are  represented  by  c. 

In  Fig.  2  let  2P  represent  the  resultant  external  force  act- 
ing on  the  left  of  AB,  both  in  amount  and  direction,  as  shown 
by  the  arrow-head  ;  then  draw  c  parallel  to  CK,  representing 
in  amount  and  direction  of  arrow-head  the  compression  in 
CK;  finally,  complete  the  quadrilateral  by  drawing  /  and  /' 
parallel  to  ED  and  CD  respectively.  The  directions  of  c  and 
2P  are  fixed  already,  hence  t  and  /'  must  have  the  directions 
indicated  by  the  arrow-heads  in  those  lines.  The  directions 
and  magnitudes  of  the  forces  /  and  f  are  thus  known,  and  H 
and  G  are  their  points  of  application.  Finally,  consider  /  and 
/'  attached  to  their  points  of  application,  as  indicated  in  Fig. 
I  ;  their  tendencies  will  be  to  lengthen  EH  and  CG  respect- 
ively, hence  ED  and  CD  will  both  be  in  tension. 

The  general  method,  then,  of  determining  the  kind  of  stress 
existing  in  any  member,  is  to  apply  to  the  point  of  division  of 
that  member  the  force  represented  in  magnitude  and  direction  by 
the  proper  side  of  the  equilibrium  polygon,  and  observe  tvhether 
the  effect  is  to  shorten  or  lengthen  the  member  in  question :  in 
the  first  case,  the  stress  will  be  compression  ;  in  the  second,  ten- 
sion. 

This  method  is  perfectly  general,  and  may  always  be  used. 

The  origin  of  moments  for  the  stress  in  CD  would  be  the 
intersection  of  CK  and  F.D  produced,  and  C  would  be  the 
origin  for  the  stress  in  ED. 


^S  NON-CONTINUOUS   TRUSSES  IN  GENERAL. 

So  far  as  the  method  is  concerned,  it  is  a  matter  of  indif- 
ference which  one  of  the  three  stresses  is  first  found  by  mo- 
ments ;  it  is  simply  advisable  to  take  that  one  which  can  be 
found  most  conveniently. 

In  all  trusses  with  vertical  loading,  '^P  is  the  algebraic  sum 
of  the  external  forces  acting  on  the  portion  of  the  truss  in 
question,  or  the  '^  shearing  stress^ 


Art.  18.— Curved  Upper  Chord— Two  Systems  of  Vertical  and  Diagonal 

Bracing. 

The  preceding  principles  will  be  first  applied  to  the  truss 
shown  in  Fig.  2,  PI.  III.  That  truss  is  taken  first,  rather  than 
a  simpler  one,  because  the  application  of  the  method  is  there 
perfectly  general,  including  all  cases  possible. 

The  truss  taken  is  composed  of  two  systems  of  triangulation, 
and  it  is  of  the  greatest  importance  to  notice  that  the  systems 
catinot  be  treated  separately,  since  the  upper  chord  is  curved, 
and  the  loads  supported  by  one  system  induce  stresses  in  the 
other.  As  an  extension  of  this  part  of  the  matter,  it  is  equally 
true  that  with  any  number  of  systems  of  triangulation,  of 
whatever  kind,  any  load  on  one  system  will  induce  stresses  in 
all  the  others.  The  result  of  this  is,  that  with  a  curved  chord 
and  more  than  one  system  of  triangulation  there  is  ahvays 
ambiguity  in  the  determination  of  stresses,  because  more  than 
three  members  will  be  severed  however  the  truss  may  be 
divided.  Certain  assumptions,  however,  may  be  made  which 
involve  no  danger,  and  which  give  a  determinate  character  to 
the  stresses  desired. 

The  assumption  in  general  terms  is  this — i.  r.,  that  at  the 
point  for  which  '^P  is  zero  the  total  load  is  divided,  and  each 
of  its  portions  travels  to  the  nearest  abutment  by  the  most 
direct  path.  This  is  a  legitimate  analysis,  so  far  as  simple 
equilibrium  is  concerned,  but  it  is  by  no  means  certain  that 
the  stresses  determined  are  those  which  actually  exist  in  the 
truss. 

In  Fig.  2  of  PI.  III.,  if  a  uniform  load  covers  the  entire 
truss  it  will  be  assumed  that  the  counters  18',  18,  19,  20,  and 


CUR  I' ED    UPPER   CHORD. 


79 


22,  and  the  correspondin<^  ones  on  the  other  side  of  the 
centre,  are  not  needed,  and  consequently  not  subjected  to 
stress. 

Again,  suppose  the  left  half  of  the  truss  to  be  loaded  with 
the  moving  load,  in  addition  to  the  fixed  load  over  the  entire 
bridge,  and  suppose  ^P  =  o  for  the  panel  point  at  the  foot 
of  diagonal  lo ;  then  it  would  be  assumed  that  all  the  verti- 
cals are  necessary,  but  only  the  diagonals  19,  20,  22,  24,  26, 
28,  30,  32,  34,  36,  and  37  in  one  part  of  the  truss,  and  10,  8, 
6,  3,  and  2  in  the  other  part. 

These  assumptions  involve  no  danger,  because  the  stresses 
deduced  by  one  legitimate  method  of  analysis  at  least  are 
provided  for.  Nevertheless  there  is  some  ambiguity,  and 
when  that  exists  there  cannot  be  economy  of  material. 

The  greatest  stresses  to  be  determined  in  the  case  treated 
will  be  found  in  accordance  with  the  preceding  principles. 

The  moving  load  will  consist  of  a  train  headed  by  four 
excesses  e,  and  will  be  supposed  brought  on  from  left  to 
right,  first  touching  the  truss  at  R. 

The  following  are  the  data  : 

Length  of  span        =  200  ft.   Height  at  centre  =     35  ft. 

Length  of  panel      =  12.5  "    Height  at  ends  =     15  " 

Number  of  panels  =   16  Radius  of  upper  chord  =  260  " 

W  (upper)  =  500  lbs.  per  foot  =  3-125  tons  /  j,.^^^  ^^^^ 

i^' (lower)  =  800    "       "       "     =5.000     "      f 

w  =  13.00  tons.  w  —  17.00  tons. 

e  ^^  w'  —  w  ■=  4.00  tons. 

The  truss  is  supposed  to  be  designed  for  a  single-track 
through  railroad  bridge. 

The  stresses  due  to  the  fixed  and  moving  loads  will  be 
found  separately,  and  those  due  to  the  fixed  load  will  be  found 
first. 

For  any  given  condition  of  loading,  the  reactions  at  the 
two  ends  R  and  R'  will  be  denoted  by  these  letters  simply. 

As  the  notation  indicates,  a  part  of  the  fixed  load  is  assigned 


8o  NON-CONTINUOUS    TRUSSES  IN  GENERAL. 

to  the  upper  chord,  according  to  the  principles  already  set: 
forth. 

For  the  fixed  load  only,  R  =  R  —  60.9375  tons  =  7.5 
{\V  +  W).  It  was  shown  above  that  the  verticals  and  main 
diagonals  are  the  only  web  members  which  will  be  assumed 
to  be  stressed  by  the  fixed  load. 

Fig.  I,  PI.  IV.,  is  the  only  diagram  necessary  for  the  deter- 
mination of  all  the  fixed-load  stresses  in  the  truss. 

The  truss  may  be  divided  through  the  members  //,  16,  and 
k  without  severing  more  than  three  members;  hence  the 
stresses  in  those  three  are  determinate. 

The  stress  in  Ji  is  the  most  convenient  one  to  find  by  the 
method  of  moments,  hence  the  middle  panel  point  of  the 
lower  chord  at  the  intersection  of  k  and  16  will  be  the  origin 
of  moments.  The  lever-arm  of  the  stress  in  h  is  found  by 
careful  measurements  (it  might  be  found  by  calculation)  on  a 
large  drawing  to  be  34.95  feet.     Taking  moments,  therefore: 

(7.5  (W^+  W)  ^  100  — 7(Jf^+  W^')  ^  50)  "^  34-95  =  93-^  tons  = 

stress  in  //. 

Hereafter,  for  the  sake  of  brevity,  the  stress  in  any  member 
will  be  represented  according  to  the  notation  of  Art.  13. 

Now  let  a  dividing  plane  cut  the  truss  in  //,  16,  and  k\  that 
portion  on  the  left  of  it  will  be  held  in  equilibrium  by  the 
external  forces  '^P^'j.^iW  ^  W)  —  ■/ {IV  +  W)  =  4.062$ 
tons  and  the  stresses  (//),  (16),  and  (k);  their  directions  being 
determined  according  to  the  preceding  principles. 

In  Fig.  I,  therefore,  of  PI.  IV.,  make  LM,  acting  upward, 
equal  to  4.0625  tons,  and  //,  acting  toward  M,  equal  to  93.1 
tons;  then  draw  k  and  16  parallel  to  the  members  indicated 
by  these  letters ;  the  directions  of  action  of  the  stresses  are 
indicated  by  the  arrow-heads,  being  the  same  around  the 
quadrilateral  in  question.  Stresses  (16)  and  {^)  are  therefore 
tension. 

The  actual  stresses  were  determined  with  a  scale  of  ten 
tons  to  the  inch  ;  but  Fig.  i  of  PI.  IV.  is  drawn  to  a  scale  of 
twenty  tons  to  the  inch. 

The  next  plane  of  division  cuts^,  16,  15,  and  k,  but  (k)  and 


CURVED    UPPER    CHORD.  8 1 

(i6)  are  known,  hence  (15)  and  {k)  may  be  determined.  For 
thisplanc,  :i^P=7.5(fF+  JF)-  -jW  -dW^UV  +  |^= 
7.1875  tons.  Lay  off  from  L  downward  a  distance  (not  let- 
tered) equal  to  7.1875,  and  from  the  lower  extremity  of  16 
draw^  parallel  to  that  panel  of  the  upper  chord  until  it  inter- 
sects LM  in  O.  As  shown,  (^)  will  act  toward  (7,  and  the 
difference  between  LO  and  7.1875  will  be  stress  (15);  it  will 
act  up,  and  consequently  will  represent  tension;  its  value  is 
1.9  tons,    g,  of  course,  represents  compression. 

The  third  plane  of  division  cuts  g,  16,  14,  and  /.  2P  = 
1{W  +  W)  =  12.1875  tons.  Hence  make  OK  equal  to 
12.1875,  ^^d  draw  14  and  /  parallel  to  those  members  ;  their 
directions  are  indicated  by  the  arrow-heads. 

The  completion  of  the  diagram  is  simply  a  repetition  of 
these  operations ;  it  is  only  necessary  to  use  care  in  giving 
the  stresses  their  proper  directions.  The  stresses  in  the 
verticals  will  be  found  in  the  vertical  line  A  T\  the  shearing 
stresses  '^P  are  also  laid  off  on  that  line,  acting  upward. 

The  following  are  the  results  of  the  complete  operation, 
+  denoting  tension,  and  —  compression: 


{h)  = 

-93.1 

tons 

{k)  = 

+  91.5 

(^)  = 

-  93-25 

(/)  = 

+  89.2 

(/)  = 

—  92.1 

(m)  = 

+  83.7 

(^)    = 

-90.4 

{n)    = 

+  78.0 

{d)  = 

-  85.8 

(0)    = 

+  66.0 

(c)    = 

-  80.5 

iP)  = 

+  497 

{b)    = 

-69.6 

{g)    = 

+  20.0 

(a)   = 

-  52.7 

(r)    = 

0 

(l)  =  -60.9375-3.125  = 


(16) 

=  + 

2.6 

tons 

(17) 

=  + 

I.O 

<< 

(15) 

=  + 

1.9 

<< 

(14) 

=  + 

4.0 

« 

(13) 

=  — 

1-5 

u 

(12) 

=  + 

8.5 

11 

(lO 

=  — 

1.4 

(( 

(10) 

=  + 

8.6 

(C 

(9) 

=  — 

6.3 

n 

(8) 

=  + 

16.5 

(( 

(7) 

=  — 

8.3 

i< 

(6) 

=  + 

21.0 

4'. 

(5) 

=  — 

I3-I 

<< 

(4) 

=  — 

20.2 

<< 

(3) 

=  + 

34-8 

(< 

(2) 

=  + 

31.8 

(( 

25  = 

-  64.0625 

(i 

•82  NON-CONTIN VOL'S    TRUSSES  IX  GENERAL. 

If  the  diagram  has  been  properly  constructed,  the  sum  of 
the  vertical  components  of  (2),  (3),  and  {a)  will  be  7.5  {W -\- 
W)  =  60.9375  tons.  In  this  case  it  came  60.25  tons,  which 
is  sufificiently  near  to  prove  the  accuracy  of  the  work.  Con- 
stant checks  may  also  be  applied  in  the  course  of  the  work, 
for  the  vertical  component  of  the  stress  in  any  diagonal  must 
be  equal  to  the  algebraic  sum  of  the  stress  in  the  vertical 
which  meets  it  at  its  foot  and  the  weight  which  hangs  from 
the  same  point. 

With  the  exception  of  a  few  web  members  near  the  centre 
of  the  truss,  the  greatest  web  stresses  will  exist  at  the  head  of 
the  train  when  it  covers  the  larger  segment  of  the  span,  pre- 
cisely as  in  trusses  with  parallel  chords.  A  number  of  the 
web  stresses,  however,  which  exist  near  the  head  of  the  train, 
for  each  of  its  positions,  will  be  given  for  the  main  diagonals, 
since  they  appear  in  the  stress  diagram  and  may  readily  be 
scaled  from  it. 

Bringing  the  moving  load  on  from  R,  panel  by  panel,  it  is 
found  by  trial  that  18'  is  the  first  counter  needed,  the  head  of 
the  train  being  at  its  foot.  For  this  position  of  the  moving 
load  R'  =  6.375  tons,  hence  the  point  for  which  2P  =  o  is  at 
the  head  of  the  train.  It  is  then  assumed  that  only  the  diago- 
nals 6,  3,  and  2  on  one  side,  and  18',  18,  19,  20,  22,  24,  26,  28, 
30,  32,  34,  36,  and  37  on  the  other  are  needed.  The  truss 
may  then  be  divided  by  a  plane  cutting  the  three  members  d, 
6,  and/  only,  and  anyone  of  these  may  be  found  by  the 
method  of  moments,  but  it  is  convenient  to  take  d.  The 
lever-arm  of  d  is  found  to  be,  by  a  large  scale,  26.7  feet. 
Hence, 

(d)  =  (R'  X  162.5)  -=-  26.7  =  —  38.8  tons. 

A  single  four-sided  stress  polygon  then  gives : 

(18)  =  +  19.00  tons. 

The  head  of  the  train  next  rests  at  the  foot  of  18.  The 
diagonals  on  the  right  of  the  head  of  the  train  which  slope 
similarly  to  18,  and  those  on  the  left  of  it  which  slope  simi- 


CURVED    UPPER    CHORD.  83 

larly  to  8,  are  needed  ;  the  others  are  not.     R  ~  10.63,  hence 
2P=  o  at  the  head  of  the  train. 

Hereafter  the  lever-arm  of  any  upper  chord  panel  will  be 
denoted  by  /,  with  the  proper  subscript.  In  the  present  case 
/g  =  29.7  feet ;  hence, 

(e)  =  R'  X  150  -^  29.7  =  —  53.7  tons, 

and  a  single  diagram  gives 

(18)  =  +  24.8  tons. 

In  order  to  save  constant  repetition,  it  may  be  stated  as  a 
general  rule  that  for  every  position  of  the  train  all  the  verti- 
cal web  members  are  needed,  but  on/j^  those  inclined  braces 
'which  slope  upzvard  and  away  from  that  panel  point  for  which 
2P  =  o  are  considered  necessary. 

With  the  head  of  the  train  at  the  foot  of  19,  R  =  15.68 
tons.  If  =  32.10  feet  ;  hence. 


Diagrams  give 


(/)  =  —  (>'].2  tons. 

(i  i)  =  +    4.0  tons. 
(19)  =  +  28.8    " 


Only  the  diagrams  for  the  next  three  positions  of  the  mov- 
ing load  will  be  given,  for  they  will  sufificiently  illustrate  the 
rest ;  they  all  embody  exactly  the  same  operations. 

Figs.  2,  3,  and  4,  of  PI.  IV.,  are  all  drawn  to  the  same  scale 
of  20  tons  to  the  inch. 

With  the  head  of  the  train  at  the  foot  of  20,  R  =21.56 
tons,  and  /^=  32.10  feet  ;  hence, 

(/)  =  {R  X  137.5  -  a'»  -^  32-10  =  -85.75  tons, 

since  '2,P  —  o  at  the  foot  of   19,  and/  is  the  panel  length. 

Divide  the  truss  through/,  19,  and  m,  then  in  Fig.  2,  PL 
IV.,   make    1  —  6  vertical,   and  f  parallel   to  that   panel  and 


84  NON-CONTINUOUS  TRUSSES  IN  GENERAL. 

equal  to  85.75  tons;  it  acts  toward  2  as  shown.  Make  2—3 
equal  to  21.56  —  ■ze'' =  4.56  tons,  and  then  draw  in  and  19. 
These  latter  must  act  in  the  direction  shown.  Now  suppose 
g,  19,  13,  and  ni  severed.  From  the  upper  extremity  of  19 
draw  ^  parallel  to  that  panel  (referring  to  Fig.  2)  until  it  in- 
tersects I  —  6  in  I  ;  I  —  2,  acting  downward,  is  the  tension 
in  13. 

It  should  be  stated  that  the  shear  2  —  3  =  4.56  tons  acts 
downward,  as  shown. 

Next  make  1—6=  21.56  tons,  acting  down,  and  draw  / 
and  20,  acting  in  the  directions  shown  ;  finally,  draw  h  and/z'. 
1  —  4,  which  acts  upward  (not  indicated),  is  the  compression 
in  15  ;  and,  in  like  manner,  4  —  5  is  the  compression  in  17. 
Scaling  from  the  diagram  : 

(11)  =  +  5.2  tons.  (20)  =  +•  16.0  tons. 

(13)  =  +  3-9     "  (15)  =  -  lo.i     " 

(17)=— 9.0     "  (19)  =  +  17.6     " 

With  the  head  of  the  train  at  the  foot  of  22,  R'  —  28.25 
tons,  and  2P=  o  at  the  foot  of  20 ;  hence  {g)  is  the  stress  to 
be  found  by  moments.     Ig  =  33.75  feet. 

{g)  =  {R'  X  125  -  w'p)  ^  33.75  =  -  98.3  tons. 

Fig.  4,  PI.  IV.,  is  the  diagram  used  for  this  position  of  the 
load,  and  is  constructed  in  precisely  the  same  manner  as  Fig. 
2.  2  —  5  is  equal  to  R'  —  w'  =  11.25  tons,  the  shear  for^,  /, 
and  20,  and  it  acts  downward,  i  —  7  is  the  shearing  stress, 
28.25  tons,  for  k,  k,  22,  and  20,  also  acting  down,  as  shown. 
I  —  2  is  the  stress  in  15  ;  2—3,  that  in  13  ;  1—4,  that  in  17 ; 
and  4  —  6,  that  in  23.     Below  are  the  numerical  values : 

(15)=+    3.8  tons.  (13)=   +   6.0  tons. 

(22)  =  +  16.0     "  (23)  =  -  10.8     " 

(20)  =  +  20.8     "  (17)  =  —  12.2     " 

Fig.  3  is  the  diagram  for  the  head  of  the  train  at  the  foot 
of  24,  and   is  constructed   in  precisely  the  same  manner  as 


CURVED    UPPER  CHORD. 


85 


Figs.  I,  2,  or  4.     R'  —  35.75,  and  '2P  —  o  for  the  toot  of  20. 
h  -  3375  feet.     Hence, 

{g)  =  {R'  X  125  -3w'/)^3375  =  -  II3-5  tons. 

(15)  =  +    4.8  tons.  (24)  =  +  25.2  tons. 

(22)=  +  15.0     "  (23)  =  -    9.4     " 

(17)  =  -    3.2     "  (25)=  -  15.7     " 

With  the  head  of  the  train  at  the  foot  of  26,  ^P  —  o  for 
the  foot  of  22,  since  R'  =  35.75  tons.    4  =  34.65  feet.     Hence, 

(//)  =  {R'  X  1 12.5  -  3w»  -^  34.65  =  -  124.7  tons. 

(22)  =  +  16. 1  tons.  (26)  =  +  34.3  tons. 
(17)=  +    7.0     "  (25)=  -    2.5     " 
(24)  =  +  12.5     "  (27)  =  —  28.0     " 

(23)  =  —  10.2     " 

With  the  head  of  the  train  at  the  foot  of  28,  R'  =  53.1875 
tons,  and  2P  =  o  at  the  foot  of  22.     4  =  34-65  feet.     Hence, 

(//)  =  {R' X  112. 5  —  6w'J>)  -^  34.65  =  —  136.1  tons. 

(17)  =  4-    6.9  tons.  (25)  =  —    1.5  tons. 

(24)=  +  12.8     "  (28)=  +25.1     " 

(23)  =  —    i.o    "  (27)  =  —  16.0    " 

(26)  =+21.5     "  (29)=  — 15.5     " 

With  the  head  of  the  train  at  the  foot  of  30,  R'  =  63.125 
tons,  and  ^P  =  o  for  the  foot  of  24.     4'  =  34-95  feet. 

.-.  (//')  z=  (R'  X  100  —  6w'p)  -^  4'  =  —  144- 1  tons. 

(17)  =  +  lo.o  tons.  {27)  =  —    9.3  tons. 

(23)  =  +    4.5     ''  (30)  =  +  35-6     '* 

(26)  =  +  15.7     "  (29)  =  —  10.7     " 

(25)=+    1.0     ''  (^31)  =-27.5     " 
(28)  =  +  20.8     ♦* 

With  the  head  of  the  train  at  the  foot  of  32,  R'  =  73-875 
tons,  and  2P  =  o  for  the  foot  of  24.     4'  =  34-95  feet. 


86 


NOX-CONTIXUOUS    TRUSSES  /.V  GENERAL. 


{h')  =  {R'  X  loo  X  low'p)  ^  4'  =  —  150.6  tons. 


(27)  =  —    9.9  tons. 

(30)  =  +  36.0    " 

(29)  =  —     I.O     " 


(32)  =  +  24.8  tons. 
(31)  =-26.8     " 

(33)  =-16.7     " 


With  the  head  of  the  train  at  the  foot  of  34,  R'  =  85.4375 
tons,  and  2P  =  o  for  the  foot  of  24.     4-  =  34.95  feet. 

.-.  (k')  =  {R'  X  100  —  i4zv'p  —  wp)  -^  34.95  =  —  154.7  tons. 

(17)  =  +    8.8  tons.  (32)  =  +  24.6  tons. 

(25)=  +    8.9     "  (31)=  -  17.4     " 

(23)  =  +    5-5     "  (34)  =  +  55-2     " 

(30)^+26.5     "  (33)  =-15.4     " 

(29)  =  -    0.8     "  (35)  =  -  46.8     " 

With  the  head  of  the  train  at  the  foot  of  36,  R'  =  97.8125 
tons,  and  ^P=  o  for  the  foot  of  24.    4'  =  34-95  feet.     Hence, 

{/i)  =  {R'  X  100  —  iSw'p  —  yup)  ^  34-95  =  —  156.5  tons. 


(17)  =  +  1 1.0  tons. 

(24)  =  —    2.0  " 
(23)  =  +    4.5  " 

(25)  =  +  12.0  " 

(27)  =  -    5-0  " 

(30)  =  +  29.8  " 

(29)  =  +    8.4  " 


(32)  =  +  12.0  tons. 
(31)  =  -20.3 

(34)  =  +  60.0 
(33)=  -  2.8 
(36)  =  +  40.0 

(35)  =  -  50.8 


With  the  head  of  the  train  at  the  foot  of  37,  R'  =  iii.oo 
tons,  and  2P=  o  for  the  foot  of  24.     4  =  34-95-     Hence, 

(/i')  =  {R'  X   icx)  —  22w'p  —  6wp)  -^  34.95  =  —  156.CXD  tons. 


(17)  r=  +  10.9  tons. 

(32)  =  +     10.5  tons. 

(24)  =   ~      2.2       " 

(31)  =  -     16.7     " 

(23)  =  +  4.4  " 

(34)  =  +     54-2     " 

(25)  =  +  II.9  " 

(36)  =  +    37-0     " 

(28)  =  +  1.5  " 

(35)  =  -    43-5     " 

(27)  =  -    5-5     " 

(37)  =  +    78.0     " 

(30)  =  +  26.0     " 

(38)=  -  1 1 1.0     " 

(29)  =  +    8.7     " 

CURVED    UPPER    CHORD. 
The  chord  stresses  with  this  loading  arc  : 


87 


(a)    —  —    87.7  tons. 

(^k)    =  +  157.6  tons 

{f)    -  -  131.2     " 

(/')    =  +  150.5      " 

{c')    =  -  137-0     " 

(;«')=    +    149-7        " 

(^^')  =  -  153-5     " 

(«•)  •-=  +  132.0    " 

(r')    =  -  152.6     " 

{0')  =  +  124.5    '' 

(/')=  -  158.7     '' 

(/)  =  +   81.8    " 

{g')=  -  156-6     " 

(g)   =  +     50.0      " 

{h')    =  -  156.0     " 

(r)   =  0 

The  same  checks  apply  as  with  the  fixed  load.  The  great- 
est stresses  are  found  by  combining  the  fixed  and  moving 
load  stresses.  The  greatest  chord  stresses  are  thus  found  to 
be  the  following : 


(^') 

= 

—  140.4  1 

:ons 

(^') 

= 

—  200.8 

(^ 

= 

-  217.5 

(''") 

= 

-  239-3 

in 

= 

--  243.0 

(/') 

= 

-  250.8 

W) 

= 

-  249.85 

(/O 

= 

-249.1 

(r)   =0 

(^')  =  +  70.0  tons. 

(/')  =  +  131-5 

(o')    =  +  190-5 

(;/')  =  +  210.0 

(;«•)  =  +  233.4 

(/')  =  +  239.7 

(k)  =  +  249.1 


In  the  case  of  the  web  stresses  the  combination  is  effected 
by  taking  the  algebraic  sum.  It  will  be  seen  that  a  number 
of  the  braces  near  the  centre  need  counterbracing,  t.  e.,  acting 
consistently  with  the  assumptions  that  were  made.  It  is 
uncertain  to  what  extent  the  existence  of  the  counters  renders 
this  counterbracing  unnecessary;  their  influence  is  therefore 
neglected. 

It  is  to  be  borne  in  mind  that  the  greatest  result  of  a  given 
sign  is  to  be  selected  from  all  the  preceding  moving-load 
stresses,  and  added,  algebraically,  to  the  stress  in  the  same- 
member  caused  by  the  fixed  load. 


88 


NON-CONTINUOUS    TRUSSES  IN  GENERAL. 


(i8')=  + 
(18)  =  + 

19.0 

24.8 

tons. 

i^7)={- 

1 1.2  tons. 
12.0  " 

(19)  =  + 

(20)  =  + 

28.8 
20.8 

(23)  =  {  - 

8.9  ' 

7-4  ' 

(22)  =  + 
(24)  =  + 

16.1 

27.8 

ii 

•(25)  =  1; 

17.2  ' 
10.5  ' 

(26)  =  + 

38.3 

i< 

(27)=     - 

29.4  ' 

(28)  -  + 
(30)  =  + 

33-6 
44.6 

(29)  =  1 ; 

21.8  ' 

2.4  ' 

(32)  =  + 

41.3 

a 

(31)=  - 

35-8  ' 

(34)  =  + 

81.0 

i(. 

(33)=  - 

29.8  ' 

(36)  -  + 

74.8 

n 

(35)-  - 

71.0  ' 

(37)  =  + 

109.8 

a 

(38)=  - 

175.0625 

tons 

In  actual  practice  it  perhaps  would  hardly  be  worth  while 
to  counterbrace  the  web  member  29. 

These  results,  then,  are  the  greatest  values  of  the  stresses 
to  which  the  different  members  of  the  truss  are  subjected. 


Art.  19. — General  Considerations. 

It  is  clear  that  in  the  graphical  treatment  of  such  a  problem 
the  stress  diagrams  should  be  as  large  as  possible.  The  scale 
used  for  all  the  results  obtained  in  Art.  18  was  ten  tons  to 
the  inch,  and  it  is  not  usually  best  to  use  more  tons  to  the 
inch  than  that. 

Another  method,  but  a  far  more  tedious  one,  of  applying 
precisely  the  same  general  principles  is  to  determine  the 
stresses  produced  in  every  member  of  the  truss  by  each  indi- 
vidual panel  load,  and  then  combine  the  results.  The  steps 
of  the  different  operations  in  such  a  case  are  precisely  the 
same  as  those  gone  through  above. 

Although  the  truss  taken  consists  of  but  two  systems  of 
triangulation,  precisely  the  same  method  is  applicable  to  any 
number  of  any  kind  of  systems,  or  to  the  ordinary  "bow- 
string" truss  of  one  system. 

In  the  latter  case  there  is  no  ambiguity  unless  counter- 
braces  are  used. 


GENERAL    CONSIDER  A  TIONS. 


89 


It  is  to  be  particularly  noticed,  also,  that  the  method  is  per- 
fectly independent  of  the  character  of  the  curve  of  the  upper 
chord  ;  it  may  equally  well  be  applied  to  trusses  with  both 
chords  curved,  or  to  trusses  with  parallel  chords  ;  in  fact,  it  is 
perfectly  general,  though  usually  not  desirable. 

Art.    20. — Position   of  Moving   Load   for   Greatest   Stress  in   any  Web 

Member. 

Two  principal  cases  occur  in  connection  with  types  of 
structures  ordinarily  used  in  engineering  practice.  That  one 
to  be  treated  first  is  the  case  of  the  intersection  of  the  chord 
sections,  in  any  panel,  lying  below  the  inclined  web  member 
in  the  same  panel ;  the  other  is  the  case  of  the  intersection 
lying  above  the  inclined  web  member.  Applications  of  these 
principal  cases  to  special  features  can  easily  be  made  after 
the  general  results  are  obtained. 

Case   1. — The  Intersection  of  Chord  Sections  below  the  Inclined  Web 

Member. 


Let  /  be  the  length  of  span  ;  i  the  distance  from  end  of 
span  to  the  point  of  intersection,  H,  of  the  chord  sections  in 
the  panel  in  question  ;  m  the  distance  from  the  end  of  the 
span  to  the  same  panel,  whose  length  is//  5  the  stress  in 
the  web  member  under  consideration,  and  h  its  lever  arm 
about  H ;  a,  b,  c,  etc.,  the  distances  separating  W^  from  Wo, 
W2  from  W^,  etc.,  etc. ;  W^,  IV^,  etc.,  the  weights  resting  be- 
tween G  and  D,  and  PV^,  JV^,  etc.,  the  weights  resting  in  the 
panel/,  while  Wn,  distant  x  from  £,  is  the  last  weight  rest- 


90  NON-CONTINUOVS    TRUSSES  IN    GENERAL. 

ing  on  the  span  from  C  to  toward  E.      b'  is  the  distance  front. 
D  to  the  nearest  weight,  W^. 

The  reaction  at  6^  is  :  • 

Ji  =  WA  ^ )+  W^\^ )+  etc...+  W„~. 

(I). 
By  taking  moments  about  H : 

Sh  =  Ri-  W,{/  +  t-a-b x)-  JV,{/  +  i-d-c x\ 

-  etc.  -  {w/-^  +  ^/  ~  ^  ~  ^  +  etc.)  {vi  +  i).     .    (2). 

By  rrioving  the  entire  load  the  distance  Ax  toward  G,  re- 
membering that  the  change  in  the  value  of  R  will  be 

A  X 

A  K  =  {Wi  +  IV2  +  •  •  •  +  IV„)  ~j  ,  the  change  in  S/i  becomes : 


{W,+  W,  +  fU.)^{m  +  t)    .     .     .     (3). 
For  a  maximum  or  minimum  a  S/i  =  o,  hence  : 


lV^  +  lV,  +  ...  +  irn=--(  IV,  +  IV,  +  etc.  )^y^W.,  +  W,^  etc.) 


pi 


(4). 


Bearing  in  mind  that  the  first  parenthesis  in  the  second 
member  of  Eq,  1^4)  represents  the  load  between  the  panel  / 
and  the  left  end  of  the  span,  and  that  the  second  represents 
the  load  in  panel  /  itself,  it  will  be  at  once  seen  that  when 
the  load  extends  from  E  to  W^,  S  is  the  maximum  main 
stress;  and  that  when  it  extends  from  G  to  lV^  {i.  e.,  to  the 
weight  farthest  toward  E),  S  is  the  maximum  counter  stress. 
Eq.  (4),  therefore,  as  it  stands,  gives  the  condition  for  maxi- 
mum main  or  counter  stress. 


GREATEST    WEB    STRESSES.  9 1 

Eq.  (4)  is  perfectly  general  in  character  and  covers  all  sys- 
terns  of  loading  whatever,  but  it  may  be  put  in  special  forms 
for  convenient  application  in  special  cases. 

Example  I. —  Uniform  Load. 

If  the  load  is  continuous,  or  only  partially  so,  and  w  is  its 
intensity  {i.e.,  its  amount  per  lineal  unit)  at  any  point  distant 
X  from  E,  then  the  various  concentrations  W^,  W^,  W^,  etc., 
will  be  represented  by  wdx.  If,  further,  the  load  is  uniform 
and  continuous,  w  is  constant,  and  W^  ■¥  W^  -\-  .  . .  +  W„  = 
wxi,  Xi  representing  the  length  of  uniform  load  on  the 
bridge.  In  the  same  manner,  if  x^  represents  the  length  of 
uniform  load  from  D  towards  G,{W]^  -\-  W„  -\-  etc.)  =  wx^, ;  and 
(W^8  +  Wi  +  etc.)  =  wrp\  r  being  the  fractional  part  of  the 
panel/  covered  by  the  uniform  load  w. 

Eq.  (4)  then  becomes  : 

/  ^  /(?«  +  i) 

wx,  =  —  ".  zvx;  +  ivrp  — : — -. 

'  I       ^  ^       pi 

Or,  x^^  -jx.i  +  rl{^^  ^  \\    ....     (5). 

As  with  the  general  case  so  with  Eq.  (5),  it  is  so  written  as 
to  give  both  maxima,  main  and  counter  stresses. 

If  the  load  is  placed  for  the  greatest  main  stress,  for  which 
x,^  —  o,  while  Xx  —  np  -\-  rp;  in  which  np  is  the  length  of  load- 
ing from  C  towards  E  : 


'b 


np  +  rp 


-/(t-)    •••    -77;;r^^  •  ■  i^\ 


\i         /      "  If. 


m 


If  the  load  is  placed  for  the  greatest  counter  stress , 


^2  =  -  7  +ir('^+i)      .     .     .     .     (7). 


92  NON-CONTINUOUS    TRUSSES  IN  GENERAL. 

Or,  as  there  is  i.o  load  between  C  and  E,  and  if  np  =  x^  is 
the  length  of  load  from  D  towards  G,  Eq.  (7)  will  become : 


\         I /         p \t         J  I (m 

Eqs.  (6)  and  (8)  will  enable  the  position  of  moving  load  to 
be  at  once  computed  without  trial. 

Example  II.  —Loads  at  Panel.  Points  Only. 

If  loads  are  located  at  the  panel  points  only,  then  JFj,  W^^, 
Ws,  etc.,  will  be  the  panel  loads,  and  a,  d,  c,  etc.,  the  panel 
lengths,  and  equal  to  each  other  in  case  those, lengths  are  uni- 
form ;  the  parenthesis  in  Eq.  (2)  multiplied  by  {in  +  /)  will 
also  disappear.  Substituting  R  from  Eq.  (i)  in  Eq.  (2)  with 
the  last  parenthesis  dropped,  there  will  result : 

Sh  —  Wxia  ^  b  +  c  +  •  "  ^  x)  (^  ^-  i  j  +  \\\  {b  +  c  +  •  •  •  ^-  x) 

(-^+i)4-.-.etc.  +  W^3(r  +  ...+;r)^+...+ff^„y' 
-(^Fi  +  ^F2  +  .-.etc.)(/  +  /) (9). 

The  last  term  of  the  second  member  represents  the  loads 
between  G  and  D. 

The  position  of  loading  for  a  maximum  of  5  will,  in  the 
general  case,  be  determined  by  trial,  by  ascertaining  at  what 
position  the  second  group  of  positive  quantities  in  the  second 
member  ceases  to  increase  more  rapidly  (as  the  load  pro- 
gresses) than  the  negative  difference  between  the  first  posi- 
tive group  and  the  negative  last  member.  This  can  only 
happen  if  the  panel  weights  toward,  or  in  the  vicinity  of 
W^  are  very  heavy  relatively  to  those  toward,  or  in  the  vicinity 
of  W^.     If  the  heavy  panel  loads  are  W-^  and  those  near  it, 


GREATEST    WEB    STRESSES.  g2a 

I.  e.,  if  the  heaviest  panel  loads  are  at  the  head  of  the  train,  the 
following  analysis  shows  the  positions  for  maxima  stresses, 
in  which,  it  is  to  be  observed,  Wi  is  the  rear  panel  load  for 
counter  stresses. 

Since  {a  +  b  +  c  -h  -  -  •  +  x)  {i  +  I)  <{i  +  1)1,  and,  hence  : 

(a  +  d  +  e  +  '  •  •  -h  ;ir)  U  +  I  j  <  (z  +  /), 

it  is  clear  that  for  maxima  main  stresses  the  loads  must  extend 

from  the  farther   ejid  of  the   span    to   the   main  tnember  in 

question. 

Since  the  counter  shear  is  negative,  i.  r.,  opposed  in  sign  to 

the  main  shear,  the  negative  portion  of  the  second  member 

of  Eq.  (9)  must  be  as  large  as  possible  for   the    maximum 

counter  shear,  and  the  positive  portion  as  small  as  possible. 

t  xt 

Hence  the  portion  W^  {c  +'•■-{•  x)-  +  ■'•  +  IV„~  must  be 

omitted,  and  the  load  W^  placed  at  the  panel  point  nearest 
the  end  G ;  i.  e.,  the  load  must  cover  that  portion  of  the  span 
between  the  eounter  and  the  nearest  end  of  the  span,  for  the 
maxima  counter  stresses. 

Hence,   for   main  web   stresses    under   the   assumed   con- 
ditions : 

S  =i^\  W-i^  {a  +  b  +  c  +  •  '  •  +  x)  +  W^  {b  +  c  ■\-  '  •  ■  -\-  x) 
+  .  .  .  +  W„x  [  ^     .     .     .     (10). 

And  for  counter  web  stresses : 
S=j\Wi{a  +  b-\-c+'-'^-x)  +  W^{b+c-\----  +  x)-ir-'-  etc.  [ 
(^+i)-(^i+  ^^.  +  ---etc.)(^'")     .     .     .     (II). 


^2b 


NON-CONTIXUOUS    TRUSSES  IN  GENERAL. 


The  conditions  on  which  Eqs.  (lo)  and  (ii)  are  based  are 

precisely  the  same  as  if  the  chords  are  parallel.     In  the  latter 

t       see  (X    \  I  -\-  t 

case  d—  oo  ,  /^  =  co  r^j  ar,  — ,  =  — ; —    -  =  o,  and  — j--  =  sec  a. 

Id         In  h 

Case  II. — The  Intersection  of  Chord  Sections  Above  the  Inclined  Web 

Member. 

This  case  is  illustrated  by  Fig.  2,  in  which  let  the  stress  S 
in  the  member  DC  be  under  consideration.    The  moving  load 


is  supposed  to  pass  on  the  bridge  from  E  towards  G  for  the 
main  web  stress.  H  is  the  point  of  intersection  of  the  chord 
sections,  while  GH  and  GD  are  the  distances  i  and  m  respect- 
ively. All  other  notation  remains  precisely  as  in  the  previous 
case.  The  reaction,  R,  under  G,  is  given  by  Eq.  (i).  Bear- 
ing in  mind  that  the  distance  DH  is  now  {i  —  ni),  and  taking 
moments  about  H,  there  will  result : 

Sh  =  Ri  -  \\\{i-l  +^  +  ^  +  "-  +  ;r)-  W^{i-l  +  <^  -h  ^  +  ••  ■  +-r) 


—  etc. 


[wjL ^'  +  W,^- ^- ""  +  etc.)  ii-m)  .   .  (12). 


Precisely  the  same  operation  which  follows  Eq.  (2)  shows 
that  the  desired  condition  for  a  maximum  is  given  by  the 
following  equation  : 


PF,  +  1^2  -(-  W^3  4-  .. .  +  W„  =  -.( W^  +  \\\  -h  etc.) 
+  (^F3+H/,-hetc.)^^'^^     .     .     .     (13). 


GREATEST    WEB    STRESSES.  g2C 

The  different  portions  of  Eq.  (13)  evidently  represent  ex- 
actly the  same  loads  as  the  same  portions  of  Eq.  (4).  It  is 
also  clear,  from  the  same  considerations,  that  if  the  loads 
extend  from  E  to  W^,  Eq.  (13)  gives  the  position  for  a  maxi- 
mum main  stress,  and  a  maximum  counter  stress  if  they  reach 
from  G  to  IV4  (/.  e\,  to  the  weight  farthest  towards  £). 

Eq.  (13),  like  Eq.  (4),  applies  to  any  system  of  loads  what- 
ever, and  can  be  applied  to  special  cases  in  the  same  manner. 

Example  III.  —  Uniform  Load. 

By  using  the  same  notation  and  the  same  process  of  reason- 
ing as  in  Ex.  i,  Eq.  (13)  takes  the  form  for  the  greatest  main 
.stress : 

Or,  for  the  greatest  counter  stress  : 

-e-o 

EXAiMPLE  IV. — Loads  at  Panel  Points  Only. 

Let  it  be  first  supposed  that  /  is  less  than  /;  i.  c,  i  <  /. 
The  same  general  considerations  that  were  given  in  Ex.  II., 
applied  to  Eq.  (12),  will  cause  it  to  take  the  form  : 

Sh^-W^{a-¥b^-c^-.'-^x){\-'^-W.Xb  ^-c  +  .'•  ^  x) 

\}  —^ij    ■  '  etc.  +  W^{c  ^  •  '  ■  ^  x)^-  -^  etc. 

+  (^^1  -H  W^a -h  etc. )(/-/)     .     .     .     (16). 
Since  {a  -{■  b  -^  c  -{■  •  •  •  ^  x){l  —  i)  <  l{l  —  i)\  hence : 

{a  +  b  +  c  +  '  ■  •  -\-  x)  (\  -yj   <{l-t). 


<^2d  NON-CONTINUOUS    TJiUSSES  IN   GENERAL. 

Therefore,  in  order  that  the  second  member  of  Eq.  (i6) 
shall  have  its  greatest  positive  value,  the  loads  must  be  at  all 
the  panel  points.  Eq.  (i6)  also  shows  that  in  the  case  now 
under  consideration  there  can  be  no  reversal  of  stress  in  any 
web  member.  In  the  inclined  member,  S,  the  stress  will 
always  be  tension,  and  always  compression  in  the  vertical 
member  passing  through  its  upper  extremity. 

The  positions  for  the  actual  maxima  stresses  can  be  found 
by  trial  only,  as  they  will  depend  on  the  amounts  of  the  panel 
loads  and  their  locations  relatively  to  each  other. 

Let  it  next  be  supposed  that  i  is  greater  than  /;  i.  e.,  i  >  /. 

In  this  case  (/ —  z)  becomes  negative,  or  (/ —  /)  positive, 
and  the  conditions  for  maxima  values  are  the  same  as  those 
fixed  for  Eq.  (9) ;  hence  they  need  no  further  attention. 

A  remaining  example  with  the  intersection  of  chord  sec- 
tions beloiv  the  inclined  web  member,  and  between  it  and  the 
end  of  the  span,  can  be  treated  in  precisely  the  same  general 
manner  as  the  preceding.  The  moving  load  between  G  and 
D  would  lie,  in  the  general  case,  partially  on  one  side  of  the 
point  of  intersection  and  partially  on  the  other.  This  form 
of  truss,  however,  has  little  or  no  technical  interest,  and  needs 
no  further  attention. 

The  preceding  treatment  applies  to  any  forms  of  trusses, 
whether  deck  or  through,  with  one  or  either  chord  horizontal. 
In  the  application  of  any  particular  formula  it  is  only  neces- 
sary that  the  point  of  intersection  of  the  chord  sections  shall 
be  located  according  to  the  conditions  on  which  the  formula 
is  based. 

The  treatment  also  applies  to  any  system  of  web  members, 
whether  they  are  all  inclined  at  different  angles  to  a  vertical 
line,  or  at  equal  angles ;  or,  again,  if,  as  in  the  Figures,  a  part 
of  them  are  vertical.  It  is  only  to  be  borne  in  mind  that  two 
web  members  intersecting  in  the  unloaded  chord  take  their 
greatest  stresses  together  only  when  that  chord  does  not  change 
its  direction  at  that  point  of  intersection.  In  case  that  direc- 
tion does  change,  the  value  of  "z"  will  be  different  for  the 
two  members,  although  "  ;;/  "  will  remain  the  same. 

The  tabulations  mentioned  in  Art.  7  and  given  in  Arts.  9 


GREATEST    WEB    STRESSES.  g2e 

and  1 1  can  be  used  in  the  application  of  the  preceding  formulae 
precisely  as  with  parallel  chords.  Short  methods  of  compu- 
tation, well  known  to  every  engineering  ofifice,  make  their 
practical  applications  easy  and  rapid. 

In  designing  trusses  with  variable  depth,  special  care  must 
be  taken  in  determining  counter  web  stresses.  It  frequently 
happens,  in  cases  similar  to  Fig.  i,  that  counters  must  be  car- 
ried at  least  one  panel  nearer  the  end  of  the  span  than  paral- 
lel chords  would  require.  Again,  the  vertical  web  members 
are  frequently  subjected  to  heavy  tension,  with  special  con- 
ditions of  moving  load,  at  panel  points  where  the  chords 
change  direction. 

Numerical  Example. 

The  truss  to  be  taken  in  this  example  is  exactly  similar  to 
that  shown  in  Fig.  i,  except  the  upper  chord  will  be  straight 
from  the  hip  to  the  top  of  the  inclined  end  posts. 

Let    /=  297  ft.  ;  /  =  27  ft.     .-.    -=ii. 

"    m  =  ip  =  ^\  ft. 

"    the  centre  depth  of  truss  be  constant  for  three 

panels  and  equal  to  42  feet. 
"    the  depth  of  truss  at  the  first  panel  point  from 

the  end  be  24  feet. 

As  the  web  member  whose  stress,  5,  is  now  desired  is  \n 
the  fourth  panel,  the  distance,  /,  is  81  feet;  /.  e.,  3/.  The 
moving  load  to  be  used  is  that  shown  in  Fig.  i  of  Art» 
y"],  and  tabulated  on  page  41.  If  this  load  passes  on  the 
bridge  from  right  to  left,  and  if  the  first  driving  wheel  rests 
at  the  panel  point  at  the  right  extremity  of  the  panel  in 
question — /.  r.,  the  fourth  panel  from  the  left  end — 93.8  feet 
of  the  uniform  load  of  3,000  pounds  per  lineal  foot  will  rest 
on  the  bridge  in  addition  to  the  two  locomotives.  In  Eq.  (4) 
the  parenthesis  (W^i  +  W^2  +  etc.)  will,  with  the  above  position 
of  loading,  be  equal  to  zero,  while  {W^  +  W^^  +  etc.)  will  equal 


92/  NON-CONTINUOUS    TRUSSES  IN   GENERAL. 

either   15,000  or  39,000,  or  some  value  between.     With  the 
values  taken  in  this  example,  Eq.  (4)  will  become : 

W^i  +  W^2  +  •  •  •  +  ^.  =  22  (W^3  +  H^4  +  .  .  •  +  etc.) ; 

and  with  the  position  of  loading  assumed,  this  equation  will 
"take  the  following  numerical  form  : 

623,400  >  (22  X  15,000  =  330,000), 
and  <  (22  X  39,000  =  858,000). 

Hence  the  position  of  loading  assumed  satisfies  Eq.  (4). 
The  resulting  reaction  at  the  left  end  {G,  Fig.  i)  of  the  span 
is  easily  and  quickly  found  by  Eq.  (i),  with  the  aid  of  the 
tabulation  on  page  41,  to  be:  7*?  =  215,000  pounds.  The 
lever  arm,  //,  of  the  stress  desired  is  found  from  the  truss 
dimensions  already  given,  and  is  15 1.2  feet.  The  distance  b' 
in  Eq.  (2)  is  18.917  feet.  The  preceding  data  substituted  in 
Eq-  (2),  gives : 

Sh  =  16,687,458     .  • .  5=1 10,370  lbs. 

The  elements  of  this  example  have  designedly  been  so 
ehosen  as  to  show  how  greatly  the  computations  may  be 
shortened  and  simplified  by  a  proper  choice  of  the  depth, 
chord  inclination,  and  intersection  distance,  /. 

Art.  21. — Position  of  Moving  Load  for  Greatest  Chord  Stresses. 

Since  the  chord  stresses  under  any  given  system  of  loading 
depend  only  on  the  truss  depths  at  the  various  panel  points, 
measured  in  a  direction  normal  to  the  opposite  panels,  the 
positions  of  loading  for  their  maxima  values  with  inclined 
•chords  will  be  identical  with  those  determined  in  Art.  7  for 
parallel  chords.  No  new  conditions,  therefore,  are  to  be 
developed  here.  The  equations  of  Art.  7  are  to  be  applied 
exactly  as  they  stand.  It  is  only  to  be  remembered  that  the 
lever  arm  for  any  panel  is  the  normal  erected  on  it  to  the 
opposite  panel  point. 


HORIZONTAL  COMPONENT  OF  GREATEST  STRESS.       93 


Art.  22. — Horizontal  Component  of  Greatest  Stress  in  any  Web  Member 
— Constant  Value  of  the  Same  for  Vertical  and  Diagonal  Bracing 
with  Parabolic  Chord — All  Loads  at  Panel  Points. 

Of  more  interest,  perhaps,  than  real  value  to  the  engineer, 
is  the  expression  for  the  horizontal  component  of  the  greatest 
stress  in  any  web  member,  though  it  may  very  easily  be 
written.     It  may  be  useful  at  times  as  a  numerical  check. 


Let  the  figure  represent  a  truss  of  one  system  of  triangula- 
tion,  subjected  to  the  action  of  vertical  loads  passing  along 
the  lower  chord  AB.  It  is  desired  to  find  the  horizontal 
component  of  the  greatest  tensile  stress  in  GH.  Let  Hh  and 
Gg  be  verticals  passing  through  //  and  G. 

The  following  notation  will  be  used  : 


n 

=  panel  length  (uniform)  in  AB. 
=  number  of  any  panel  from  B ; 
2,  and  4  for  OH. 

for  PQ, 

n  has 

the 

value 

rp 

d 

d' 

=  Mg 
=  Gg 
=  Hh. 

N 

=  number  of  panels  in 

AB. 

I 

w 

R 

=  AB=:Np. 

=  moving  panel  load. 

=  reaction  at  B. 

nip 

=  BM. 

For  the  greatest  tension  in  GH  the  moving   load    must 
extend  from  B  to  H. 


94  NON-CONTINUOUS    TRUSSES  IN  GENERAL. 

The  distance  of  the  centre  of  such  a  load  from  B  is  — i^. 

2 

Hence, 

^=(/--f-)7(«.-.)-  =  -|('^.-.).-2i^>^| 

Now  let  moments  be  taken  about  G. 
Hence, 

(J///)  =  {r(„,  -  r)p  -  («,  -  ,)  a.  I  («,  _  ^)^  -  M  I  ^^  ^ 

Or,  (MH)  =  ^  [^-<y-)^J  .,  («,  -  .)    .     (.). 

In  order  to  obtain  the  horizontal  component  of  the  stress 
in  GF,  due  to  the  assumed  load,  it  is  only  necessary  to  take 
moments  about  H  in  precisely  the  same  manner.  The  ex- 
pression, however,  can  be  derived  immediately  from  Eq.  (i) 
by  putting  r  =  i,  and  writing  <r/'  for  d. 

.-.  Hon  Com.  {GF)  =  j^  [^  -  (^^i  -  0 /]  ^^  (;,^  _i)   .    (2). 
2a  I 

The  horizontal  component  of  the  greatest  tensile  stress  in 
GH  is  the  difference  between  the  second  members  of  Eqs.  (i) 
and  (2) ;  let  it  be  called  //j. 

...  i/,  =  ^f  I  '-(^^'-'^p  -  ^-<y>^  I «,  („,  - .) .  (3). 

If  a  is  the  angle  of  inclination  of  GH  to  a  horizontal  line^ 
then : 

{GH)  =  H^sec  a (4). 

Eqs.  (3)  and  (4)  apply  to  all  tensile  web  stresses.  For  com- 
pressive web  stresses  as  typified  by  {GM)  there  would  be 
found  the  Hon  Comp.  {GK),  instead  of  Hon  Comp.  (GF), 
by  taking  moments  about  A/;  d'  would  then  represent  Mm. 
By  making  r  =  o  in  Eq.  i : 


hORIZOXl-AL    COMPONENT  OF  GREATEST  STRESS.       95 
Hor.  Comp.  {GK)  =  ^|;  ^^ - '^^P^   n,  [u,  -  i)    .     (5). 

Hence,  for  the  horizontal  component  of  the  greatest  com- 
pressive stress  in  GM : 

And,  {GM)  =  H,' sec  a'     ....    (7). 

By  means  of  the  Eqs.  (3),  (4),  (6),  and  (7),  every  web  stress 
in  the  truss  may  be  determined  by  formula. 

If  GM  is  vertical,  r  =  o  and  d  =  d'  m  Eq.  (6),  and  ///  =  o, 
as  it  should. 

If  GH  is  vertical,  Eq.  (3)  shows  H^  to  be  zero  in  the 
same  manner. 

It  is  to  be  borne  in  mind,  in  the  application  of  these  formu- 
las, that  n  is  counted  along  the  loaded  segment ;  also  that 
d',  for  tension,  is  taken  at  the  head  of  the  train,  and  one 
panel  in  front  of  it  for  compression. 

If  the  moving  load  passes  along  the  upper  chord,  exactly 
the  same  formulae  hold  true,  but  d'  taken  at  the  head  of  the 
train  will  give  compression,  and  tension  when  taken  a  panel 
length  in  front. 

If  the  curve  KFC  is  a  parabola,  with  vertex  at  the  centre 
of  the  span,  if  K  and  C  coincide  with  A  and  B,  respectively, 
and  if  GM  and  all  corresponding  web  members  are  vertical, 
Eq.  (3)  becomes : 

H,  =  i|^  I  '^.TLll/  _  LlA^l }  „,  („,-.).  (8). 

From  the  ordinary  equation  to  the  parabola : 

j^  =  ax 

and  —  =  adi ; 

4 


96  NON-COXTIXUOUS    TRUSSES  IX  GEXERAL. 

in  which  d-^  is  the  depth  of  the  truss  at  the  midle  of  the  span. 
Hence, 

^  =  ^37"- 

In  this  equation  put  y  — r^p  and  x  =  d^  —  d,  then  j  = 

('^1  —  0/  and  X  =  d^  —  d',  successively.     There  will  re- 
sult: 

4  4"i 

4  4«i     . 

Remembering  that  /  =  Np : 

4  («i  TV  —  «i^) 


^=4 


JV^ 


Putting  these  values  in  Eq.  (8),  also  Np  =  /,  there  will  re> 
suit: 

^  _  wfN^  (         TV-  (;?i-  I)  N-n^      { 

'  -  ~Mjr  \  K-  I)  N-{n,-i)~  n^N-n^  \  "'  ^'''       '^ 

As  this  is  the  horizontal  component  of  the  greatest  tension 
in  any  diagonal  web  member  and  constant,  t/iaf  greatest  stress 
itself  is  the  hypothejmse,  parallel  to  the  brace  in  question,  of  a 

right-angled  triangle  of  wJiich  the  base  is  H-^  —  — — . 

8^4 


B01VS7'R/.VG    7 A' CSS.  gj 

This  furnishes  a  very  short  method  of  finding  the  stress  in 
any  inclined  web  member. 

The  similarity  between  //j  and  the  total  stress  in  the  hori- 
zontal chord,  with  the  truss  wholly  loaded,  is  interesting. 

If  the  trussing  is  so  designed  that  the  diagonal  or  inclined 
braces  sustain  compression,  Eq.  (6)  gives  precisely  the  same 
general  result,  but  with  the  sign  changed. 

In  such  a  case  there  would  be  substituted  in  the  parabolic 

/                                                               / 
equation  y  = «,/  and  x  —  d-^  —  d'  also,  r  = {n^  —  i)/ 

and  X  —  d^  -  d\  <■/  and  rt''  having  changed  places. 

If  no  web  members  are  vertical,  y  will  have  for  one  value 

/ 

in   the   equation    to    the   parabola,    -  —  [n^  —  r) p   instead   of 

^hP^  the   other  values  to   be   substituted  remaining  the 

same.     This  new  value  gives, 


^i{n,-r)N-{_n,-rn 


Now  making  the  substitutions  in  Eq.  (3)  instead  of  Eq.  (8) 


8^1  \«i  —  I       n^  —  rj    ^^  ^         ' 


"^  =  Td,  (^)  "• 


Art.  23. — Bowstring  Truss — Diagonal  Bracing — IjKample. 

The  first  form  of  bowstring  truss  to  be  treated  is  that  shown 
in  Fig.  I.  All  braces  are  inclined,  and  each  apex  in  the  upper 
chord  is  vertically  over  the  centre  of  the  panel  below. 

The  truss  is  supposed  to  be  designed  for  a  highway  bridge^ 
There  is  a  sidewalk  on  either  side. 


98 


NON-CON TINUOU S    TRUSSES  IN  GENERAL. 


The  greatest  moving  load  will  be  assumed  to  be  that  of  an 
advancing  crowd  of  people,  from  the  left  end  of  the  span, 
weighing  eighty-five  pounds  per  square  foot. 

As  the  span  is  a  short  one,  and  the  roadway  heavy,  the 
whole  of  the  fixed  load  will  be  put  upon  the  lower  chord. 

The  following  are  the  data  required  : 

Span  =  72  feet.     Depth  of  truss  at  centre  =11.7  feet. 
Radius  of  circumference  of  circle  passing  through  apices  in 
upper  chord  =  60  feet. 

Number  of  panels  =  6.     Panel  length  =  12  feet. 

Width  of  roadway,  from  centre  to  centre  of  trusses  =  20  feet. 

Width  of  each  sidewalk  =  6  feet. 

\V  =  900  pounds  per  foot  =  5.4  tons  per  panel. 

%v  =  32  X  85  X  12  =  16.32  tons  per  panel. 

■As  usual,  W^and  w  refer  to  fixed  an3  moving  loads  respect- 
ively. 


C    -101.0      2> 


Fig.  I. 


In  all  the  diagrams  that  follow,  the  lines  indicated  by  any 
two  letters  are  parallel  to  the  members  of  the  truss  at  the 
extremities  of  which  the  same  letters  are  found. 

In  this  truss  and  in  the  two  which  follow,  the  upper  and 
lower  chord  sections  found  in  any  panel  intersect  outside  of 
the  span  RR ,  hence  the  positions  of  the  moving  load,  for  the 
greatest  web  stresses,  are  precisely  the  same  as  those  which 
would  be  taken  for  a  truss  with  parallel  chords. 

With  the  head  of  the  moving  load  at  J/,  the  truss  is  first 
to  be  considered  as  divided  through  the  members  AB,  BM, 
and  ML  ;  then  through  BC,  BL,  and  ML. 


BOWSTRING    TRUSS. 
Fig.  2  is  the  complete  diagram  for  this  position. 


99 


R  (reaction)  =  27.1  tons. 

{ML)  ={R  X  18  —  21.72  X  6)  -^  9.3  =  38.4  tons.     9.3  feet  is 
the  depth  of  truss  through  B. 


«  ore; 


Fig.  2. 


The  shear  is, 


5  =  27.1  —  21.72  =  5.38  tons. 

The  diagram  needs  no  explanation.     It  gives: 

{BM)  =  +  18.5  tons.  {BL)  =  -  2.1  tons. 

With  the  head  of  the  moving  load  at  L:  R  =  37.98  tons. 
Fig.  3  is  the  complete  diagram. 


(CD) 


Fig.  3. 

The  truss  is  first  supposed  to  be  divided  through  BC,  CL^ 
and  LK\  then,  through  CD,  CK,  and  LK. 

{LK)  =  {R  X  30  —  2  X  21,72  X  12)  ^-  1 1.7  =  52.8  tons. 

The  shear  is, 

-S"  =  37-98  -  43-44  =  -  546  tons. 


lOO 


NON-CONTINUOUS    TRUSSES  IN  GENERAL. 


The  diagram  gives: 

{LC)  —  +  20.0  tons. 


{CK) 


0.60  tons. 


With  the  head  of  the  moving  load  at  AT:  R  —  46.14  tons. 
Hence, 

{KH)  =  (7?  X  42  —  3  X  21.72  X  18)  H-  1 1.7  =  65.4  tons. 
Fig.  4  is  the  complete  diagram. 


(CD) 


W» 


Fig.  4. 

The  shear  is, 

5  =  46.14  —  65.16  =  —  19.02  tons. 

The  diagram  gives: 

(KD)  =  +  21.7  tons.  (^//)  =  -  7.4  tons. 

With  the  head  of  the  moving  load  at  //:  R'  =  40.7  tons, 
and  S  =  -  (40.7  -  5.4)  =  -  35.3  tons. 

{BG)  =  {R'  X  18  -  5.4  X  6)  -^  9.3  =  75.3  tons. 


mE) 


Fig.  5. 
Fig.  5  is  the  diagram,  and  it  gives : 

(//£)  =  +  20.9  tons.  i-^G)  =  —  4.0  tons. 


ROW  STRING   TRUSS. 


lOI 


With  the  head  of  the  moving  load  at  G,  or  with  the  moving 
load  over  the  whole  bridge  :  7?  =  2.5  x  {W  +  w^^  54.3  tons. 
In  this  case  no  chord  stress  is  found  by  moments,  but  the 
diagram,  Fig.  6,  is  worked  up  from  the  end  of  the  truss. 


(KT^) 


Fig.  6. 


It  gives: 

{RA)  =  —  100. 1  tons. 
(^^)  =-109.1     " 
\bC)  =  -  103.0     " 
\CD)  =  -  102.0     " 


{RM)  =  +  84.6  tons. 
\mL)   =  +  93.6     " 
{LK)  =  +  97.0     - 
{AM)  =  +  19.5     " 


The  results  of  these  diagrams  are  collected  and  written  in 
Fig.  I. 

Both  web  and  chord  stresses  may  be  checked  by  moments 
as  follows. 

Moments  about  K  give  : 

{CD)  =  —  (54.3  X  36—2  X  21.72  X  18)  -f-  1 1.7  =  —  100.2  tons. 

The  diagram  gave  102.0  tons.  The  agreement  is  close 
enough  for  the  purpose,  but  in  an  actual  truss  the  difference 
ought  not  to  be  greater  than  one  per  cent,  of  the  smallest 
result. 

Again,  BC  and  LK  intersect  in  a  point  about  30.8  feet  to 
the  left  of  R,  and  the  normal  distance  from  that  point  to  CL 


102 


nojv-coa't/xl'ous  ti^usses  in  general. 


produced  is  about  48.5  feet.  Hence,  with  the  head  of  the 
moving  load  at  L,  and,  by  taking  moments  ^bout  the  point  of 
intersection  : 

{CL)  =  (2  X  21.72  X  48.8  -  7?  X  30.8)  ^  48.5  =  +  19.6  tons. 

The  diagram  gave  +  20.0  tons.  The  agreement  is  suf- 
ficiently close. 

Numbers  of  checks  like  the  two  above  should  be  applied. 

The  Fig.  i  shows  that  the  web  members  MB,  BL,  CL,  CK, 
DK,  DH,  HE,  and  EG  must  be  counterbraced. 


Art.  24. — Bowstring  Truss — Vertical  and  Diagonal  Bracing  with 
Counters. 

The  case  next  to  be  taken  is  that  of  an  ordinary  "  bow- 
string" truss  with  vertical  and  diagonal  bracing,  and  is  rep- 
resented in  Fig.  I.  The  inclined  braces  are  not  supposed 
capable  of  resistance  to  a  compressive  stress.  Inasmuch  as 
counters  are  almost  invariably  introduced  in  such  a  truss  in 
ordinary  engineering  practice,  they  will  be  supposed  to  exist 
in  this  case. 


Fig.  I. 

The  truss  is  supposed  to  be  designed  for  a  highway  bridge, 
furnished  with  sidewalk  on  each  side. 

The  greatest  moving  load  will  be  taken  as  a  crowd  of  peo- 
ple weighing  eighty-five  pounds  per  square  foot,  covering 
roadway  and  sidewalks,  advancing  panel  by  panel  from  R 
until  the  truss  is  entirely  covered. 

Since  the  span  is  a  short  one,  and  the  roadway  very  heavy, 
the  whole  of  the  fixed  load  will  be  taken  as  applied  to  the 
lower  chord. 


BOirs  TRIXG    1  'R  USS. 


103 


The  following  are  the  data  required  ;  they  are  taken  from 
the  preceding  Article; 

Span  =  72  feet.  Depth  of  truss,  12  feet. 

Radius  of  circumference  of  circle  passing  through  upper  ex- 
tremities of  verticals  —  60  feet. 
Number  of  panels  =  6.  Panel  length  =  12  feet. 

Width  of  roadway,  from  centre  to  centre  of  trusses  =  20  feet. 
Width  of  each  sidewalk  =  6  feet. 
W  =  900  pounds  per  foot  =  5.4  tons  per  panel, 
w  =  32  X  85  X  12  =  32,640  pounds  =  16.32  tons  per  panel. 

As  in  Art.  18,  if  the  plane  dividing  the  truss  cut  more  than 
three  members,  some  one  of  these  members  must  be  neg- 
lected or  assumed  not  to  exist. 

For  the  sake  of  brevity,  two  letters  inclosed  by  a  paren- 
thesis will  denote  the  stress  in  the  member  indicated  by  these 
letters. 

The  placing  the  load  for  the  greatest  web  stresses,  is  done 
in  accordance  with  the  general  principles  established  in  Art. 
20. 

When  the  head  of  the  moving   load  is  at  Z,  the  existence 

of    AK  must    be    ig-  «^ (j^jq 

nored  ;  for  if  BL  be 
then  omitted,  y^  A' will 
suffer  compression. 

With  the  head  of 
the  train  at  Z,  R—Zj.i 
tons ;  hence 


Fig,  2. 


{LK)  =  (/?  X  24  -  21.72  X  12)  ^  {BK=  10.8)  =  36.1  tons. 

Fig.  2  is  the  complete  diagram  for  this  case,  and  explains 
itself,  ac  is  the  shear  '^P  —  27.1  —  21.72  =  5.38  tons.  Scaling 
from  Fig.  2  : 

{BL)—  +  19.2  tons.  {BK)  —  +  2.0  tons. 

BH  is  also  omitted  for  this  loading. 


104 


NO.V-CONTINUOUS  TRUSSES  IN   GENERAL. 


With  the  head  of  the  moving  load   at   K,  R  =  37.98  tons, 
and  CG  and  BH  are  omitted.    {KH)  —  [Ry.  36—2  x  2 1 .72  x 


18)  -^{CH=  12)  =  48.78  tons.  Fig.  3  is  the  diagram  for  this 
case,  ce  is  the  shear,  acting  downwards,  2P^=  37-98  —  2  x 
21.72  =  —  5.46  ton. 

This  diagram  gives  the  results  : 


iCH) 


i.o  ton. 


(AT)  =  +  16.0  tons. 


With  the  head  of  the  moving  load  at  H,  R  =  46.14  tons. 
DF  and  CG  will  be  omitted. 

It  is  unnecessary  to  give  the  diagram  for  this  case, 
drawn  precisely  as  the  two  preceding  ones  have  been. 

The  diagram  gives : 


It  is 


(DG)  =  -1.0  ton. 


(HD)  =  +  1 7. 1  tons. 


Neither  will  the  diagram  for  the  head  of  the  load  at  G  be 
given,  as  it  is  constructed  exactly  like  the  others.  It  gives, 
omitting  GC  and  DF: 

{GE)—  +  15.00  tons. 

For  the  greatest  chord  stresses  the  moving  load  covers  the 
entire  truss,  and  Fig.  4  of  the  next  Article  is  the  complete 
diagram  for  the  case.  All  the  explanation  which  the  diagram 
needs  is  there  given.  BL,  KC,  GC  and  FD  are  supposed  to 
be  omitted. 

Taking  moments  about  B,  with  uniform  load  {iv  +   W)  • 

{KH)  =  (54.3  X  24  —  21.72  X  12)  ^  10.8  =  96.5  tons. 

Others  may  be  checked  in  the  same  way. 
The  most  rational  circumstances  under  which  the  greatest 
tensile  stresses  can  be  supposed  to  occur  in   the  verticals,  are 


BOWSTRING   TRUSS. 


105 


those  under  which  they  are  found  in  the  next  Article  ;  and 
the  results  there  obtained  are  used  in  this  case.  They  are 
introduced,  without  more  explanation,  in  Fig.  i.  Their  dia- 
grams will  be  found  in  the  next  Article. 

These  results  are  by  no  means  satisfactory,  but  nothing 
better  can  be  done  with  such  a  form  of  truss. 

Some  of  the  web  stresses  should  be  checked  by  moments, 
by  the  general  method. 

Fig.  I  shows  the  greatest  web  stresses  selected  from  all  the 
preceding  results. 

It  is  far  more  convenient  to  treat  the  fixed  and  moving 
loads  together,  as  has  been  done  in  this  case,  than  to  treat 
them  separately,  as,  of  course,  may  be  done. 

Again,  the  stresses  caused  by  each  panel  load  on  all  the 
members  of  the  truss  may  be  found,  and  their  effects  com- 
bined, but  this  also  requires  far  more  labor  than  the  method 
followed. 

The  stress  diagram  for  each  position  of  the  moving  load 
might  have  been  worked  up  from  the  end  of  the  truss,  as  was 
that  for  the  chord  stresses,  but  it  saves  considerable  labor  to 
find  one  chord  stress  by  moments,  and  begin  the  diagram 
with  that. 


Art.  25.  —  Bowstring    Truss  —  Vertical   and  Diagonal    Bracing    without 

Counters. 

It  is  evident,  from  what  has  preceded,  that  the  existence  of 
the  counters  causes  considerable  ambiguity  in  the  web  stresses, 


Fig.  I. 


and  it   is  much  more    satisfactory    from   a   strictly  technical 
point  of  view  to  leave  them  out,  as  shown  in  Fig.  i. 


io6 


NON-CONTINUOUS    TRUSSES  IN  GENERAL. 


Fig.  I  is  exactly  the  same  as  Fig.  i  of  the  preceding  Arti- 
cle, with  the  counters  omitted,  and  in  this  Article  will  be 
found  the  stresses  existing  in  it  with  precisely  the  same  data 
as  were  used  above. 

The  moving  load  is  brought  on  panel  by  panel  from  R,  ac- 
cording to  the  general  principles  established  in  a  preceding 
Article. 

With  the  head  of  the  moving  load  2X  L,  R  =  27.1  tons.  As 
before,  {LK )  =  46.9  tons. 


(HP 


Fig.  2. 
Fig.  2  is  the  complete   diagram  for  this  loading.     {RL)  — 
{LK )  =  46.9  tons,  and  ^<^  =  27.  i  —  2 1 .72  =  5.38  tons. 

Hence,  {^^)  =  ii-9  tons. 

With  the  head  of  the  moving  load  at  K,  R  =  37.98  tons. 
{KH)  =  (R  X  24  —  12  X  21.72)  -i-  10.8  =  60.27  tons. 

Fig.  3  is  the  diagram  for  this  loading,  and  it  explains  itself. 
Hence, 


(BK)  =  +  24.9  tons. 


{BH)=  -  15.8  tons. 


Fig.  3. 

It  is  unnecessary  to  show  the  diagrams  for  the  two  cases  of 
the  head  of  the  moving  load  at  H  and  G.  They  give  respect- 
ively : 


BOWSTRING    TRUSS.  10/ 

{DH)  =  +  17.1  tons,  {DG)=  —  i.o  ton,  and  {GE)  =  +  15.0  tons. 

Fig.  4  is  the  diagram  for  the  moving  load  over  the  whole 
bridge,  bd  is  the  reaction  R  =  R'  =  54.30  tons  ;  dc  =  21.^2 
tons  ;  dixxd  fg  is  equal  to  \  (21,72)  tons. 

The  greatest  chord  stresses,  taken  from  Fig.  4,  are  written 
in  Fig.  I. 

The  greatest  web  stresses  are  selected  from  all  the  preced- 
ing results,  and  also  written  in  Fig.  i. 

The  stress  {BC)  may  easily  be  checked  by  moments,  as 
follows,  by  taking  the  origin  at  H.  The  normal  distance  of 
H  from  {BC)  is  1 1;9  feet.     Hence, 

{BC)  =  ;2^(zt'+  ^F)x36-3x  \2{zu+W)\  -m  1.9=98.56  tons. 

Other  and  similar  checks  should  also  be  applied. 

It  is  seen  in  Fig.  i  that  the  diagonals  need  counterbracing, 
but  there  is  no  ambiguity,  and  the  superiority  of  the  design 
over  that  in  the  preceding  Article  is  evident. 


(RI^) 


2fh  =  (EC)  -  -naeton^i 


Fig.  4. 


It  is  to  be  carefully  borne  in  mind  that  the  diagrams  must 
always  be  drawn  as  large  and  as  accurately  as  possible.  Those 
of  the  present  Article  were  constructed  to  a  scale  of  ten  tons 
to  the  inch.     The  figures  do  not  show  the  scale. 


io8 


NON-CONTINUOUS    TRUSSES   IN    GENERAL. 


Art.  26. — Deck  Truss  with  Curved  Lower  Chord,  Concave  Downward 
— Loads  at  Panel  Points — Exsunple. 

The  truss  shown  in  Fig.  i  is,  in  some  respects,  a  peculiar 
one.  It  has  one  prominent  characteristic  which  distinguishes 
it  from  the  bowstring  trusses  which  have  been  treated  in  the 
three  preceding  articles,  in  that  the  chord  sections  (upper  and 
lower)  in  any  panel,  excepting  those  two  at  the  centre,  inter- 
sect in  the  upper  chord  within  the  limits  of  the  span.  All 
the  web  stresses,  except  those  in  Z?iV  and  DM,  will  have  their 
greatest  values  when  the  moving  load  covers  the  whole  truss, 
as  was  shown  in  Art.  20.     NJM  is  horizontal,  consequently  the 


JL     -3».»      X      -  sa.9    C 


Fig.  I. 

intersections  of  NM  with  CD  and  DE  are  found  at  an  infinite 
distance  from  the  truss,  but  ON  and  CD  intersect  between  E 
and  7^  (near  the  latter  point) :  all  other  intersections  are  found 
between  A  and  G. 

The  positions  of  moving  load  for  the  greatest  stresses  in 
DN  and  DM  are  the  same,  therefore,  as  for  a  truss  with 
parallel  chords. 

Observations  relating  to  the  positions  of  the  points  of  inter- 
section of  the  chord  sections  in  the  figure  apply  to  a  truss  for 
which  the  following  are  the  data: 

Radius  of  circumference  of  circle  passing  through  the  apices 
of  the  lower  chord  =  60  feet. 

Vertical  distance  of  centre  of  circle  below  D  =  66  feet. 
Depth  of  truss  chrough  D  =  6.3  feet. 
Span  =  AG  =  y2  feet. 
AR  =  GR     =  18     " 


DECK    TRUSS. 

Uniform  upper  chord  panel  length  —  12  feet. 
NM  =  12  feet. 

ON  =  ML  =  IT,      " 
OP  =  LK  =  1 1.7  " 

Uniform  panel  fixed  load       =    5.4  tons  =  W. 
"  ''      moving  load  =  16.32  '^     =  w. 


109 


(CS) 


(BT) 


CDir)/,\ 

/             \          / 

-.  /    j\ (SMI 

-T \ —     y 

f  * — 

\IP) 

/     \/ 

•  \     4 

^       fsc;       _V 

c 

\. 

«lX   =2?   =i'«.5   ivrx^ 

jl,M«j_V 

Ic   ^cd  ^  21.  jz 

d-e  =  (C2); 

fc  ^  r.MTv-; 

Fig.  2. 

The  loading  is  the  same  as  that  used  in  the  preceding  bow- 
string trusses. 

Let  the  angle  which  DN  or  DM  makes  with  a  vertical  line 
be  denoted  by  ot.     Then, 

tan  ix.  —  0.952  ;  sec  (v  —  1.38. 

For  greatest  tensile  stress  in  DN. 
With  the  moving  load  extending  from  A  to  C\ 

Reaction    R  —  37.98  tons. 
The  shear  5  =  2  x  21.72  —  37.98  =  5.46  tons. 
Hence,  {^^)  —  S  sec  a  =  +  7.53  tons. 

For  greatest  compressive  stress  in  DM. 
With  the  moving  load  extending  from  A  to  D: 

Reaction    R  =  46.14  tons. 

The  shear  5  =  46. 14  —  3  x  21.72  =  —  19.02  tons. 
Hence,  (DM)  =  S  sec  a  =  —  26.25  tons. 


no 


NQN-COA'TIN UO U S    TRUSSES  IN  GENERAL. 


For  the  other  web  stresses  the  moving  load  must  cover  the 
whole  truss,  and  Fig.  2  is  the  complete  diagram  for  that  con- 
dition of  loading.  With  the  data  given  below  the  figure,  no 
explanation  is  needed.  The  results  of  the  diagram  will  be 
found  in  Fig.  i,  together  with  those  determined  above  by  the 
trigonometrical  method. 

One  advantage  inherent  in  this  form  of  truss,  as  in  all  in 
which  the  chord  intersections  are  found  within  the  limits  of 
the  span,  is  the  little  counterbracing  required. 

In  the  truss  taken,  the  two  web  members  DN  and  DM  dive 
all  that  require  such  treatment.  Indeed,  with  a  sufificiently 
small  radius  of  lower  chord  and  centre  depth,  together  with 
an  odd  number  of  upper  chord  panels,  a  truss  may  readily  be 
designed  which  will  require  no  counterbracing  at  all.  A  dis- 
advantage, however,  is  the  small  depth  at  centre,  just  where 
a  great  one  is  needed,  with  the  resulting  heavy  chord  stresses. 

As  the  Fig.  i  shows,  no  stress  exists  in  PR  or  KR ;  never- 
theless those  members  would  ordinarily  be  inserted  for  the 
purpose  of  stiffening  the  whole  structure. 

The  truss  may  be  supported  directly,  as  at  G,  or  there  may 
be  an  end  post,  as  AR. 

The  greatest  stress  in  AR,  will  be  the  reaction  R  added  to 
^{w  -\-  W).     Hence, 

{AR)  =  —  (54.3  +  10.86)  =  —  65.16  tons. 

As  checks,  moments  about  D  give : 
(NAI)  =  (54.3  X  36  —  2  X  21.72  X  18)  -^  6.3  =  +  186.0  tons. 

The  normal  distance  from  C  to  ON  is  7.5  feet  (nearly). 
Hence,  by  moments  about  C: 

(ON)  —  (54.3  X  24  —  21.72  X  12)  -^  7.5  =  +  139.0  tons. 

OP,  prolonged,  cuts  the  upper  chord  at  a  point  about  three 
feet  from  D  toward  £,  and  the  normal  distance  from  that 
point  of  intersection  to  OB,  prolonged,  is  about  23.5  feet. 
Hence, 

(OB)  =  (54.3  X  39  -  21.72  X  27)  ^  23.5  =  +65.1  tons. 


CRANE    TRUSSES. 


Ill 


The  agreement  of  the  last  result  with  that  obtained  by 
diagram  is  very  close,  but  the  other  results,  by  the  two 
methods,  show  a  difference  of  about  two  per  cent.  This  is 
close  enough  for  the  present  purpose,  but  in  practice  the 
figure  and  diagrams  should  be  drawn  large  enough  to  make 
this  difference,  at  most,  one  per  cent,  of  the  smallest  result. 

A  truss  of  this  character,  with  more  than  one  system  of 
triangulation,  gives  indeterminate  stresses,  but  the  approxi- 
mate method  of  Art.  i8  may  be  used.  Approximate  deter- 
minations may  also  be  made  by  treating  each  system,  with  its 
weights,  by  the  methods  just  given,  and  combining  the  results 
for  the  chords. 


Art.   27. — Crane    Trusses. 
A  form  of  truss  which  has  been  used   for  powerful  cranes, 

A 


Jl.^ 


Fig.  I. 


1 1 2  ^^0A^-  CO  A'  TIN  UO  US    TR I  \SSE  S  IN   GENERA  I. 

under  circumstances  requiring  much  head  room,  is  that  shown 


m  Fig.  I.     It  revolves  about  a  vertical  axis  midway  between 
E  and  F. 


CRANE    TRUSSES.  H^ 

In  the  example  taken,  the  weight,  W,  hanging  from  D,  the 
peak,  is  supposed  to  be  ten  tons. 

Each  chord  of  the  truss  ;«,  /,  k,  etc.,  or  a,  b,  c,  etc.,  is  made 
up  of  chords  of  quadrants  of  two  circumferences  of  circles. 
The  radius  for  the  chord  mlk,  etc.,  is  25  feet,  and  that  for 
the  other  chord  is  22.6  feet.     EF  is  5  feet. 

Denoting  the  chord  panels  by  single  letters: 

rt'  =  5  feet. 
m  =  b  =  c  =  6    " 
/  =  ./=;    " 

/i  =  g   " 

Fig.  2  is  the  complete  diagram  for  the  stresses  in  the  truss, 
supposing  the  on/j/  load  to  be  the  ten  tons  hanging  from  the 
peak.  If  it  should  be  necessary  to  take  into  account  the 
weight  of  the  truss,  it  would  be  done  precisely  as  the  fixed 
weights  of  trusses  have  been  treated  in  the  preceding  Articles. 

The  lines  in  Fig.  2,  denoted  by  letters  and  figures,  are  par- 
allel to  lines  denoted  by  the  same  letters  and  figures  in  Fig.  i. 

The  diagram  gives  the  following  results : 


(0  = 

+  12.5  tons. 

(  2)  = 

—  14  tons. 

(3)  = 

+  6.00  " 

(4)  = 

-  13.7  " 

(5)  = 

+  1.50  " 

(6)  = 

-  134  " 

(7)  = 

-  2.60  " 

(  8)  = 

—  14.0  " 

(9)  = 

—  7.20  " 

(10)  = 

-  13-2  " 

(II)- 

-  5-' 

8  tons. 

{a)  = 

—  8.7  tons. 

(.«)  = 

+  17.8  tons. 

{b)=^ 

—  24.0  " 

(/)  = 

+  30.4  " 

i^)  = 

-  35-9  " 

{k)  = 

+  38.7  " 

{d)  = 

-44.0  " 

{h)  = 

+  44-2  " 

(^)  = 

-48.2  " 

{g)  = 

+  47-2  " 

(/)  = 

-  54 

. I  tons. 

These  results  may  easily  be  checked  by  moments.      The 


114 


NO^r-CONTINUOUS  TRUSSES  IN  GENERAL. 


different  lever  arms,  with  two  exceptions,  to  be  used,  are 
shown  in  Fig.  i.  The  normal  distance  from  ^  to  C  is  about 
4.6,  and  from  e  to  B  about  5.3  feet.  These  lever  arms  were 
scaled  from  the  drawings,  and  may  not  be  exactly  right,  but 
near  enough  for  the  purpose.     By  moments  about  C\ 

{S)  =  +  (10  X  22.25)  ^  4-6  =  +  48.4  tons. 

By  moments  about  B\ 

{e)  ~  —  (10  X  26.0)  H-  53  =  —  49.0  tons. 

The  chord  sections  d  and  k,  prolonged,  meet  at  ^,  and  mo- 
ments about  that  point  give  : 

(7)=  -  (10  X  3.7) 


14.7 


2.5  tons. 


These  results  agree  sufficiently  well  with  those  obtained  by 
the  diagram. 

If  the  chain,  rope,  or  cable  pass  along  either  chord,  the  ten- 
sion in  it  will  tend  to  produce  an  equal  amount  of  compres- 
sion in  the  panels  of  that  chord.  The  resultant  stress,  there- 
fore, in  any  panel  will  be  the  algebraic  sum  of  this  amount  of 
compression,  and  the  stress  due  to  the  weight  W. 

Fig.  3  is  a  skeleton  diagram  of  the  ordinary  crane,  which 
revolves  about  the  centre  line  of  ED  as  an  axis.  AB  is  the 
--^  weight  hung  at  the  peak,  A.  BC 
is  parallel  to  DA,  and  represents 
the  amount  of  tension  in  that 
member.  AC  \?>  the  compression 
in  AE  due  to  the  weight  W.  As 
before,  the  tension  in  the  rope  or 
cable  tends  to  produce  an  equal 
amount  of  compression  in  any 
member  along  which  it  lies. 

Let  /  denote  the  normal  dis- 
tance from  DA  to  any  point  in 
the  centre  line  of  DE\  then  any 


Fio.  3. 


section  of  DE  will  be  subjected  to  the  bending  moment 
3/=  {DA)  X  /. 


ROOF   TRUSSES. 


II.S 


DE  will  also  be  subjected  to  a  direct  stress  (tension  in 
Pig.  3)  equal  to  the  vertical  component  of  the  stress  in  DA. 
The  greatest  resultant  intensity  of  stress  in  any  section  will 
be  the  combination  of  the  intensities  due  to  these  two  causes. 

Art.  28. — Preliminary  to  the  Treatment  of  Roof  Trusses — Wind  Pressure 

— Notation. 

Four,  only,  of  the  principal  types  of  roof  trusses  with 
straight  rafters  will  be  treated,  since  the  method  used  for 
any  one  is  precisely  the  same  in  character  as  that  to  be  used 
for  any  other. 

The  wind  will  be  assumed  to  act  on  one  side  of  the  roof, 
and  its  resultant  action  will  be  assumed  to  be  normal  in  direc- 
tion to  the  rafters.  If  such  is  not  the  case,  f  and  i>  will 
represent  empirical  determinations  of  the  horizontal  and 
vertical  components,  respectively,  in  the  Articles  which  fol- 
low, and  the  methods  will  remain  precisely  the  same. 

Let/  be  the  intensity  of  this  normal  wind  pressure,  and  let 
/  and  /i  be  the  lengths  of  two  adjacent  panels  of  the  rafter, 
while  ^  is, the  horizontal  distance  between  two  adjacent  and 
parallel  rafters.  The  total  normal  wind  pressure  supposed 
exerted  or  concentrated  at  the  point  between  the  two  panels, 

will  then  be  — — ^^ ~  ;  or  if  /  =  /i ,  as  is  usually  the  case,  pdl. 

If  ^  is  the  angle  which  the  rafter  makes  with  a  horizontal 
line,  the  horizontal  component  of  this  normal  pressure  will  be : 

/=  ^"^^^^  ^^)  sin  6  ■  or  pdl  sin  6  • 

and  the  vertical  component  will  be  : 

pd[/  +  /i)  ^  ,,         „ 

V  =  ~ ^ cos  6  ;   or  pdl  cos  0. 

Finally,  let  the  total  fixed  weight  of  the  roof  be  supposed 
concentrated  at  the  panel  points  of  the  rafters ;  and  let  the 
weight  of  such  load  at  any  panel  point  be  represented  by  W^. 
The  total  vertical  load  at  any  panel  point  will  then  be  : 

The  vertical  reactions  due  to  the  vertical  component  of  the 


Il6  A'ON-CO.VTIXUOUS    TRUSSES  IN   GENERAL. 

wind  pressure  and  the  fixed  load  are  found  by  the  principle 
of  the  lever  in  the  usual  manner  ;  that  at  the  left  of  the  span 
will  be  called  R,  and  that  at  the  right,  R' . 

The  vertical  reactions  due  to  the  horizontal  forces  /,  will, 
however,  be  called  R\  and  they  will  be  found  for  the  different 
cases  by  taking  moments  about  any  point  in  the  horizontal 
line  joining  the  feet  of  the  rafters.  If  b  is  the  vertical  pro- 
jection of  a  rafter,  and  2c  the  span,  or  distance  between  the 
feet  of  two  rafters  meeting  at  the  ridge,  there  will  result : 

^         2        2C  4C 

At  the  foot  of  the  rafter  pressed  by  the  zvind,  this  reaction 
will  be  downward  in  direction  ;  at  the  foot  of  the  other  rafter 
it  will  be  uptvard. 

The  total  horizontal  reaction  at  the  points  of  support  will 
be  equal  to  bp,  the  total  horizontal  force  of  the  wind,  and  its 
direction  of  action  will  be  opposite  to  that  of  the  wind.  The 
horizontal  component  of  the  wind  pressure  and  the  horizontal 
reaction  produce  a  couple,  equal  and  opposite  to  that  whose 
force  is  R",  and  whose  lever  arm  is  2c. 

R"  must  be  numerically  less  than  R,  or  the  wind  will  turn 
over  the  roof  bodily. 

If  the  foot  of  one  rafter  is  supported  on  rollers,  the  hori- 
zontal reaction  will  be  wholly  exerted  at  the  foot  of  the  other. 

If  the  foot  of  neither  rafter  is  supported  on  rollers,  the 
horizontal  reaction  will  be  assumed  to  be  equally  divided 
between  the  points  of  support. 

The  stresses  for  the  vertical  and  horizontal  loads  are  found 
by  separate  diagrams,  although  they  might  be  found  by  one 
only,  because  the  slope  of  the  roof  may,  in  some  cases,  be  so 
small  as  to  make  it  needless  to  consider  the  forces/". 

If  rollers  are  used  at  the  foot  of  one  rafter,  the  wind  may 
press  that  one  or  the  other.  In  treating  a  large  roof  it  may, 
then,  be  necessar\^  to  take  the  wind  first  in  one  direction  and 
then  in  the  other. 

These  two,  with  the  case  of  no  rollers,  make  three  possible 
cases,  and  an  example  will  be  taken  in  each  one. 


ROOF    TRUSSES. 


117 


Art.  29.— First  Example. 

The  truss  represented  in  Fig.  i  is  a  roof  truss,  applicable 
to  short  spans.  There  is  no  "  moving  load  "  in  such  a  case. 
The  wind  pressure,  however,  may  act  on  one  side  and  not  on 
the  other,  and  for  that  reason  VV,  \\\,  and  W^^  are  taken  as 
differing  from  each  other,  as  was  explained  in  the  preceding 
Article. 


tiM^ 


Fig.  I. 

There  is  no  essential  error  in  this  case  in  assuming  all  the 
load  concentrated  at  the  points  indicated.  The  wind  press- 
ure is  assumed  to  act  on  the  left  side  of  the  truss,  so  that 
W>W^>  W^.  Also,  JFi  =  1  ( JF+  W^).  A  and  B  are  equal 
in  length,  and  E  is  horizontal. 

Figs.  2  and  3  are  the  stress  diagrams  for  Fig.  i,  and  the  lines 
in  them,  indicated  by  letters  primed,  are  parallel  to,  and  repre- 
sent stresses  in  the  members  marked  with  the  same  letters  in 


Fig.  2. 


Fig.  I.  The  kinds  of  stresses  in  the  different  members  are 
shown  by  the  signs  +  or  — ,  in  both  figures,  signifying  tension 
or  compression,  respectively. 


Il8  NOA'-CONTINUOUS    TRUSSES  IN  GENERAL. 

Fig.  2  is  the  diagram  for  the  vertical  loads,  and  Fig.  3  that 
for  the  horizontal  loads/". 


{^ 

c^__^ 

-7^ 

^ 

-S-    ^ 

<r^ 

^ 

VI.'    /T 

b" 

K 

■< 

-B. 

>^ 

yi 

^ 

" 

< 

^-i^ 

-^ 

^ 

^^ 

^--/,    -X  -;^     -X-;S 


Fig.  3. 

Rollers  are  supposed  to  be  under  the  foot  of  neither  rafter. 
Consequently  the  horizontal  reaction  at  each  end  is  /,  as 
shown,    /i  is  equal  to  hf- 

In  Fig.  I,  R"  acts  downward  at  the  left  of  the  span,  and 
upward  at  the  right. 

The  resultant  stress  in  any  member  is  the  algebraic  sum 
of  those  given  by  the  two  diagrams. 

Fig.  3  does  not  show  all  its  lines  parallel  to  the  members  of 
Fig.  I.  There^  is,  of  course,  a  diagram  similar  to  Fig.  2,  for 
the  right  half  of  the  truss,  but  it  is  not  needed. 

The  stresses  may  also  be  found  by  the  method  of  moments, 
by  locating  the  origin  of  moments  according  to  the  general 
principle  stated  in  Art.  17. 


Art.  30. —  Second  Example. 

Fig.  I  of  this  Article  represents  a  very  common  roof  truss. 
As  before,  the  total  load  is  supposed  concentrated  on  the 
rafters  at  the  points  indicated.  Wind  pressure  is  taken  as 
acting  on  the  left  of  the  roof,  making  JV >  JJ\  >  JV.2.  Jl\  is 
simply  the  panel  weight  of  the  roof,  and  U\  =  }^{IV+  W^. 
Each  rafter  is  divided  into  four  equal  parts,  and  W  is  taken 
equal  to  2W^_.  Hence  the  reaction  R  —  3  fF.  This  does  not 
at  all  affect  the  generality  of  the  diagram.     The  lower  ex- 


KOOF    TRUSSES. 


119 


tremity,  however,  of  D',  in  Fig.  2,  will  not  usually  be  found 
at  a'. 


Fig.  I. 

Fig.  2  is  the  stress  diagram  for  the  vertical  loading  taken, 
and  the  notation  has  precisely  the  same  meaning  as  before. 

It  is  seen  from  Fig.  i,  that  there  is  some  ambiguity  in 
regard  to  the  stresses  C,  Q,  F' ,  and  M'.  It  may  be  assumed, 
however,    that   E'  =  H'   and   P'  =  Q',   which    makes   F'  = 


Fig.  2. 


The  kinds  of  stresses  are  shown  on  the  diagrams. 


I20 


NOJV-COA^TIXUOUS    T/^USSES  IX  GENERAL. 


Fig.  3    is  the   diagram  for  the   horizontal   stresses  f  and 

Rollers  are  supposed  to  be  placed  at  the  foot  of  A.  Hence 
all  horizontal  reaction  will  be  found  at  the  foot  of  A^.  As 
shown,  that  reaction  will  be  equal  to  ^f.     The  vertical  reac- 


tion R"  will  be  directed   downward   at  the  foot  of  A,  and 
upward  at  the  foot  of  A-^. 

Considering  the  horizontal  forces  only : 

Ai  =Bi  =  c  =  n,' =  -A'. 
K,'=L,'=0'  =  -4/- A". 

The  resultant  stress  in  any  member  is  found  by  combining 
the  results  of  the  two  diagrams  in  the  usual  manner. 

This  style  of  truss  is  so  frequently  used  that  formulae  for 
the  stresses  due  to  the  vertical  loading  are  given  below,  in 
which  a  is  the  length  of  the  rafter,  c  the  half  span  RR\  and  d 
the  height  of  W^  above  O.  These  expressions  may  be  readily 
derived  from  Fig.  i. 


A'=.jR, 


B'  =A'-W 


C  =  A'  -  2lV~, 
a 


D'  =  A'  -iW 


H'=E'  =-W. 
a 


F'  =  2H'  =  2-W, 
a 


ROOF    TRUSSES. 


121 


0  ~  '00 

O'  =L'  -Jlf'=j{R-^lK),       N'  =  M'  +  S'  =  |^  ^' 


Art.  31.— Third  Example. 

Figs.  I,  2,  and  3  are  roof-truss  and  stress  diagrams,  respect- 
ively. The  wind  pressure  is  supposed  to  act  on  RIV^,  and 
the  total  load  is  taken  to  be  concentrated  as  shown. 


-/^   v3 


sp 


Fig.  I. 


cte  ^  /^' 

act,  R 
dc  Z1C6  zw 


Fig.  2. 


The  rafters  are  each  divided  into  three  equal  parts. 


^22 


NON-CONTINUOUS    TRUSSES  IN  GENERAL. 


The  notation  has  the  same  signification  as  that  used  be- 
fore, and  it  is  unnecessary  to  explain  the  diagrams. 

Fig.  2  is  the  diagram  for  the  vertical  loading,  and  Fig.  3, 
that  for  the  horizontal  forces/" and /i  =  \f. 

Rollers  are  supposed  to  be  placed  at  the  foot  of  A-^  in  Fig. 
I.  Hence  the  total  horizontal  reaction  will  exist  at  the  foot 
of  A,  and  its  value  will  be  3/",  as  shown.  As  before,  the 
direction  of  R"  will  be  upward  at  the  foot  of  A-^,  and  down- 
ward at  the  foot  of  A. 

The  resultant  stress  in  any  member  will  be  found  by  com- 
bining the  results  given  in  Figs.  2  and  3. 


\--^'^-~\\ 

^'>^ 

'.C',' 

^ 

^  f 

<  f     H-~. 

'x'^-+ 

r 

Fig.  3. 

Fig.  3  does  not  show  K'  equal  to  ///,  as  it  is  not  a  scale 
diagram. 

The  members  FH  and  K  are  in  tension,  while  D  and  E  are 
in  compression.  Wy  Wx>  W.^,  the  last  being  the  fixed  panel 
weight  of  the  roof. 

Art.  32. — Fourth  Example. 

In  this  Article,  as  before,  Fig.  i  is  the  truss,  and  Fig.  2  is 
its  stress  diagram  for  the  vertical  loading.     Wind  pressure  is 


Fig.  I. 


GENERAL    CONSIDER  A  TIONS. 


123 


taken  as  acting  on  RW]^\  the  reaction  R,  therefore,  is  \  (3  W  + 
\\\  +  W.^  and  W>  \\\>  W^.  The  rafters  are  each  divided 
into  three  equal  parts,  and  F  and  D  are  vertical.  IV2  is  the 
fixed  panel  weight  of  the  roof,  and  W^  =  ( IV  +  W.2)  -i-  2. 


Fig.  2. 


The  kinds  of  stresses  in  the  various  members  are  shown  in 
Fig.  2. 

It  is  unnecessary  to  give  the  diagram  for  the  horizontal 
components  of  the  wind  pressure.  The  method  of  drawing 
it  will  fall  under  some  one  of  the  three  cases  already  given. 


Art.  33. — General  Considerations. 


In  all  the  preceding  cases  of  roof  trusses  the  stresses  may 
be  obtained  by  the  method  of  moments,  and  therefore  the 
graphical  results  may  be  checked  by  that  method. 

The  operations  are  precisely  similar  if  a  part  of  the  load  is 
hung  from  points  in  the  tie  RR',  in  all  the  cases,  or  if  the 
loading  is  even  more  eccentric  than  that  assumed. 

In  many  cases  the  sides  of  large  buildings  are  braced  to 


124  NON-CONTINUOUS    TRUSSES  IN  GENERAL. 

the  roof  by  oblique  members  extending  from  that  panel  point 
of  RR'  adjacent  to  the  side,  to  some  point  in  that  side.  In 
such  cases  the  wind  will  cause  stresses,  in  the  different  mem- 
bers of  the  roof  truss,  which  must  be  determined  independ- 
ently of  those  already  found  and  added,  algebraically,  to 
them. 


CHAPTER    IV. 

SWING   BRIDGES.    ENDS   SIMPLY    RESTING   ON   SUPPORTS. 
Art.  34. — General  Considerations. 

Without  regarding  the  nature  of  the  supports  or  attach- 
ments at  the  extremities  of  the  two  arms,  swing  bridges  are 
divided  into  two  classes :  those  with  center-bearing  turn- 
tables, and  those  with  rim-bearing  turn-tables.  In  the  first 
class  the  entire  reaction  at  the  pivot  pier  is  exerted  through 
a  central  pivot,  or  a  nest  of  one  or  more  series  of  solid,  con- 
ical rollers ;  usually  the  latter.  In  such  a  case  there  may  be 
a  circular  drum  or  framework,  supported  on  wheels  running 
on  a  circular  track,  but  they  are  used  solely  for  the  purpose 
of  steadying  the  bridge  while  open. 

In  the  case  of  a  rim-bearing  turn-table,  however,  the  reac- 
tion at  the  pivot  pier  is  exerted  through  the  circular  track  on 
which  the  wheels  supporting  the  drum  or  framework  of  the 
table  turn.  The  object  of  a  pivot,  in  such  a  case,  is  simply 
to  enable  the  bridge  to  turn  truly  about  a  center. 

In  reference  to  the  truss,  there  is  evidently  only  one  point 
■of  support,  at  the  pivot  pier,  with  a  center-bearing  turn-table, 
i.  c,  at  the  center. 

With  a  rim-bearing  turn-table,  however,  there  may  be  two 
or  more  points  of  support  at  the  pivot  pier,  though  it  will  be 
shown  hereafter  that,  by  separating  the  different  systems  of 
triangulation,  it  will  never  be  necessary  to  consider  more  than 
two  at  once. 

Again,  with  either  turn-table  there  may  be  three  different 
methods  of  supporting  or  securing  the  extremities  of  the  two 
arms  of  the  bridge.  These  extremities  may  simply  rest  on 
.supports,  so  that  the  reaction  will  always  be  zero,  or  upward  ; 

125 


126  SWING  BRIDGES. 

this  is  the  only  case  which  will  receive  more  than  a  passing 
notice  in  this  chapter. 

Again,  those  extremities  may  be  fastened  down  or  latched 
to  the  piers  when  the  bridge  is  not  open.  The  reaction  may 
then  be  nothing,  upward  or  downward. 

In  these  two  cases  the  reactions  at  the  extremities  of  the 
arms  will  be  zero  when  the  bridge  is  simply  closed,  and  sup- 
porting no  moving  load. 

In  the  first,  when  the  moving  load  is  on  one  arm,  the  ex- 
tremity of  the  other  may  be  slightly  raised  from  its  support  ; 
in  the  second  case,  however,  that  extremity  will  be  held  down 
by  the  latching  apparatus,  /.  e.,  the  reaction  will  be  downward. 
The  object  of  the  latching  apparatus  is  thus  seen  to  be  the 
prevention  of  thchammering  of  a  truss-end  on  its  support. 

Finally,  the  third  method  is  to  raise,  by  proper  machinery, 
the  truss  ends,  when  the  bridge  is  closed,  any  diesired  amount. 

The  object  of  this  arrangement  is  to  insure  a  reaction  at 
the  extremities,  which  will  always  be  nothing  or  upward,  and 
thus  obviate  the  hammering  before  mentioned. 

In  this  case  the  whole  of  the  bridge  weight  does  not  rest  at 
the  pivot  pier,  as  the  lifting  of  the  ends  takes  up  a  part  of  it; 
in  fact,  may  take  up  the  whole  of  it. 

Recapitulating,  then,  the  ends  of  a  swing  bridge  may  be : 

(l.)  Simply  supported, 

(2.)  Latched  down, 

(3.)  Lifted  up. 

The  detailed  consideration  of  the  first  case  will  next  be 
taken  up. 

Art.  35. — General  Formula  for  the  Case  of  Ends  Simply  Supported— Two 
Points  of  Support  at  Pivot  Pier — One  Point  of  Support  at  Pivot 
Pier. 

With  two  points  of  support  at  the  pivot  pier  there  usually 
arises  the  case  of  a  continuous  beam  resting  on  four  points  of 

v4 ^^  :b      l^     C £5 j> 

^ —  i^  ^  a 

2d,  M^  M-^  M^ 


GENERAL  FORMULA.  12/ 

support,  as  shown  in  Fig.  i.  The  notations  of  the  spans  and 
bending  moments  at  the  different  points  of  support  are  suf- 
ficiently well  shown  in  the  figure.  The  points  of  support 
will  all  be  taken  in  the  same  horizontal  line,  as  the  formulae 
will  then  also  apply  to  any  configuration  belonging  to  a  state 
of  no  stress,  provided  the  truss  may  be  considered  straight 
between  any  two  points  of  support  (see  Appendix).  Any 
truss  may  be  considered  straight  when  an  equivalent  solid 
beam  has  a  neutral  surface  which  is  plane  before  flexure  ; 
a  straight  solid  beam  is  "  equivalent  "  to  a  straight  truss  when 
equal  moments  of  inertia  and  resistance  are  found  at  the  same 
section  in  the  two  structures. 

The  theorem  of  three  moments,  in  the  ordinary  form,  does 
not  apply,  then,  to  a  continuous  truss  with  one  chord  curved, 
and  none  of  the  following  investigations  apply  to  such  a 
case. 

Again,  in  the  span  /,  there  will  be  supposed  no  load,  as 
such  is  usually  the  case.  The  load  on  /,  ought  always  to  be 
supported  on  short  girders  or  beams  resting  at  B  and  C,  for 
there  is  the  less  complication  of  stresses  in  the  trusses,  and 
consequently  less  liability  to  uncertainties  ;  besides,  such  an 
arrangement  is  probably  more  economical  in  material. 

In  the  present  case  M^  and  M^  will  each  be  equal  to 
zero. 

Let  z  denote  the  distance  of  the  point  of  application  of  any 
force,  P,  from  the  left-hand  end  of  the  left-hand  span,  or 
right-hand  end  of  right-hand  span.  In  /j,  z  would  be  meas- 
ured from  A,  while  in  4  it  would  be  measured  from  D. 

The  formula  expressing  the  theorem  of  three  moments  for 
all  supporting  points  in  the  same  level  becomes,  by  using  the 
notation  of  Fig.  i  : 

M4x  +  2Mlk  +  4)  +  M4.  +  yi/^/i'  -  ^)s  +  \hp{li  -  s')z  =  o. 

1  2 

The  symbols  2  and  2  indicate  summations  for  the  spans 
/i  and  4. 

Applying  the  above  equation  to  spans  4  and  4,   and  then 


128  SWING  BRIDGES. 

to  4  and  4>  there  will  result,  bearing  in  mind  the  circumstances^ 
of  the  present  case : 

2J/2(/,  +  4)  +  J/3/2 +  7i/'(/i'-^)^  =  0     .     (i). 

M^  k  +  2J/3  (4  +  4)  -f  yhp{ii  -z')z  =  o    .    (2). 

If  Equation  (i)  be  multiplied  by  /„  and  Equation  (2)  by 
2  (4  +  4) ;  and  if  the  results  so  obtained  be  subtracted,  and 
if  the  following  notation  be  used  : 

4,4  2  s 

'2  '2  *i  '3 

there  will  result : 

,^       c'l^hPii  -n')n-  2(^4  (r  +  I)  ^P{i  -  ;«2)  ;«         ,  , 

Equation  (i)  then  gives  : 

_  M^  c%^P{i  -n')n  ,. 

'  2(^+1)  ...     V4;- 

Equation   (2)    will    evidently   give    another   expression    for- 
M2,  but  it  is  not  necessary  to  write  it. 

Let  Rx,  R2,  Rsj  and  R^  be  the  reactions  at  the  points  of 
support  A,  B,  C,  and  D  respectively,  Figure  i.  Then,  adapt- 
ing the  formulae  for  reactions  from  the  theorem  of  three  mo- 
ments (see  Appendix)  to  the  notation  of  the  present  case,, 
there  may  at  once  be  written 

R,=  :SP(i-n)+^ (5). 

R,^2P»-^-^'-^'      .    .     .     (6). 

4  4 

^8=  +  ^-^ +^^^--/-    •    .    (7)- 

R,=  2P{i  -;«)  +-^       ....     (8).. 

4 

As  should  be  the  case,  there  is  found : 

R,  +  R,  +  R,  +  R,  =  2P  +  2P. 


GENERAL   FORMULA.  1 29 

In  ordinary  swing  bridges  where  1^  =  1^  =  /,  and  a  single 
weight  P  rests  on  /,  : 

R.=p{^-ri)\'~^'^''^''~- — r~^l   •  (5-). 

R,  =  c\R,-R,-P{i-7i)\-R,     .     .     {7a). 
R,:=P-Ri-  Ri  +  R,  .     .     .     .     (Sa). 

It  may  sometimes  be  convenient  to  use  the  following  equa- 
tion derived  from  Equation  (i): 

J/3  =  -  ;  2 J/2  (r  +  I)  +  r/2^P{i  -  ;/2)  n\      .     (9). 

These  are  all  the  equations  necessary  for  the  solution  of 
the  case  of  two  supports  at  the  pivot  pier,  frequently  exist- 
ing if  the  turn-table  is  rim-bearing. 

If  there  is  only  cme  point  of  support  at  the  pivot  pier,  the 

-A  I, S z^  c 

e. zs- ^ 

M,  :m:^  M^ 

case  reduces  to  that  of  a  continuous  beam  of  two  spans  only, 
as  shown  in  Figure  2. 

As  A  and  C  are  points  of  support  only,  M^  and  J/3  are  each 

zero;  hence  if  m  now  stands  for  y,  the  equation  immediately 

preceding  Equation  (i)  gives: 

^.  =  -//i^'-'^)"^^f^' -'-')"'■  .  (.0). 

-»  (<r  +  l)  ^     ' 


There  will  also  result 

R,=  ^P(^i-n)+^^ (II). 


J.  ^,  M, 


R^=2Pn-^+iPm-^  .     .     .     (12). 

/l  la 


1-1 


-  M 

R,  =  2P{i-m)+'^     ....     (13). 
'2 


^30  SWING  BRIDGET. 

If  l^=  i.^  =  /,  then  c  =  I  and  Equation  (lo)  becomes : 
M,  =  --   I  2P{i  -  n')  n  +  ip(i  -  ;;r') ;«  '        .      .      (13^). 


There  is  again  found  : 


1 


R^  +  R..  +  Rs=  2P  +  2P. 

These  complete  the  general  formulae  needed  for  the  case  of 
ends  supported. 

Some  very  important  deductions  are  to  be  drawn  from 
Eqs.  (5),  (6),  (7),  (8),  (11),  (12),  and  (13),  considering  them 
applied  to  bridges  with  rim-bearing  ttirn-tables. 

Those  equations  are  so  written  that  a  positive  value  of  R 

means  a  reaction  upward  in  direction,  while  a  negative  value 

indicates  a  downward  reaction. 

In  the  case  of  Fig.   i,  let  the  span  4  be  supposed  free  of 

3 
any  loads  P,  then  the  term  involving  the 'summation   '^  will 

disappear.  Eq.  (3)  then  shows  that  M^  will  always  be  posi- 
tive ;  consequently  Eq.  (4)  shows  that  M^  will  always  be 
negative. 

Using  these  results  in  connection  with  Eqs.  (5),  (6),  (7),  and 
(8),  it  is  at  once  seen  that  R^  and  R^  ivill  always  be  positive, 
while  R^  will  ahvays  be  negative. 

It  may  also  be  shown  that  Eq.  (5)  makes  R^  positive  in  such 
cases  as  always  arise  in  an  engineer's  practice,  although  that 
equation  apparently  shows  that  R-^  may,  under  some  circum- 
stances, be  negative,  since  M^  is  always  negative  in  the  case 
taken,  while  (/,  —  z)  is,  of  course,  always  positive. 

By  deducing  the  value  of  M.-^  from  Eqs.  (3)  and  (4),  and 
introducing  it  in  Eq.  (5),  there  will  result: 


^■4 


"vP  r/  -  -^  -  1  i  ^  (^^  +  ^^  \ 

_       ^'     ^^      /,  \4  (/1  +  I.)  (I.  +  4)  -  4V 


4  \4  (/i  +  4)  (4  +  4)  -  /i 

ip(4— ^-)- 

If  4  is  very  small  in  comparison  with  4  or  4,  and  if,  at  the 


GENERAL   FORMULAE. 


I  ".I 


same  time,  z  is  small,  there  may  be  written  for  a  single  "/*," 
nearly : 

I 


R,= 


/: 


~Ph-iij  .  A^^]; 


which  is  evidently  positive. 

If,  on  the  contrary,  L  is  small  and  z  large,  there  may  be 
written  for  a  single  weight  P,  nearly : 


n 


p  (A  -  z)  -^p{i,^  z)  ik  -  z)  r\. 


In  this  expression  R^  can  be  equal  to  zero  only  by  suppos- 
ing the  negative  quantity  within  the  brackets  to  be  larger, 
numerically,  than  it  ought  to  be,  /.  c,  by  making  (/j  +  z)  =  21^, 
and  z  =^  l^;  hence  it  can  never  be  negative. 

\i  li=z  l^=l^  =  I  the  general  value  of  R^  takes  the  form  : 


R,= 


k 


_P(l-z)-^^^P{l'-z^)?\. 


If  z  is  small,  there  results  nearly 


R,= 


Pl-^P'. 
15 


If  z  is  large,  nearly 


^■  =  /^ 


PU-  £)-—P 
15 


1 

{l-z)\ 


In  neither  case,  therefore,  can  the  reaction  be  negative. 

It  may,  consequently,  be  assumed  as  a  principle  that  if  h  is 
small  in  reference  to  k  or  4,  or  if  it  is  equal  to  those  quan- 
tities, the  reaction  R\,  in  the  case  supposed,  must  always  be 
positive ;  and  within  those  limits  must  be  found  all  cases  of 
swing  bridges. 

Since  any  load  on  the  span  /j  makes  the  reactions  at  A  and 
D  positive,  considered  by  itself,  so  any  load  on  4  will,  of 


132 


SWING  BRIDGES. 


itself,  make  the  reactions  at  D  and^  positive.  Consequently, 
as  there  is  never  any  load  on  4,  the  reactions  at  A  and  D  wilt 
always  be  positive,  and  the  ends  of  the  bridge  will  never  tend  to 
rise  from  their  points  of  support.  No  "hammering,"  there- 
fore, can  take  place,  in  this  case,  at  the  ends. 

The  case  of  the  two  points  of  support,  B  and  C,  taken  in 
connection  with  a  center-bearing  turn-table  will  be  considered 
further  on. 

Fig.  2  represents  the  case  of  either  a  center  or  rim-bear- 
ing turn-table  with  only  one  point  of  support  at  the  pivot 
pier;  the  two  cases  are  coincident  in  all  their  circumstances. 

Eq.  (lo)  shows  Mo  to  be  always  negative.  Consequently,  if 
there  is  no  load  on  4,  R-^  and  R.^  will  always  be  positive,  while 
7?3  will  always  be  negative. 

When  span  /j  carries  load,  however,  the  span  4  may,  at 
the  same  time,  support  just  enough  load  to  make  R^  equal 
to  zero;  more  load  than  that  will  make  R^  positive,  or  upward 
in  direction. 

But  in  the  present  case  the  point  C  is  simply  a  point  of 
support,  consequently  no  negative  or  downward  reaction  can 
exist  there.  It  becomes  necessary,  therefore,  to  determine 
just  how  much  load  on  4»  combined  with  the  full  load  on  4* 
will  make  R^  =  o.  For  this  purpose,  M^  must  be  taken 
from  Eq.  (lo)  and  inserted  in  Eq.  (13),  while  in  the  latter  R^ 
must    be   just    equal  to  zero.     For   the    sake  of  brevity,  let 

-  '^P  (4'  —  ^')  s  be  represented  by  A. 
4 

Then  from  Eq,  (10)  : 

^  is  a  constant  quantity  so  far  as  this  operation  is  con- 
cerned. 

Putting  this  value  of  M^  in  Eq.  (13)  after  making  R^  =  o: 

2l,{l,  +  l^hp{h-z)-^P{l.?-^^z  =  AL  .     (15). 
,    Eq.  (15)  indicates  what  disposition  of  the  ioad  on  the  span 


ENDS  SIMPLY  SUPPORTED.  j,^ 

4  will  make  -^3=0;  and  it  will  be  seen  to  be  of  very  easy 
application. 

The  determinations  of  7?i,  R^,  and  R^,  for  all  loading  in 
excess  of  that  indicated  by  Eq.  (15)  will  require  the  use  of 
Eqs.  (11),  (12),  and  (13)  as  they  stand;  for  all  loading  less 
than  that  amount,  however,  R.^  —  o,  and  the  reactions  R^  and 
^2  are  to  be  found  by  the  simple  principle  of  the  lever,  con- 
sidering l-i  as  a  simple  overhanging  arm,  or  cantilever.  These 
operations  will  be  shown  in  detail  hereafter. 

In  this  case  it  is  evident  that  "hammering  "  will  take  place 
at  the  ends  with  certain  dispositions  of  loading. 

Art.  36. — Ends  Simply  Supported — Two  Points  of  Support  at  Center — 
Partial  Continuity — Example. 

The  general  formulae  of  the  preceding  Article  will  be 
applied  to  the  truss  shown  by  skeleton*  diagram  in  Figure  i  of 
Plate  X,  in  which  the  intermediate  verticals /"a  and  Pg,  with  CP 
and  the  inclined  web  members  EP  and  /\ ,  are  in  compression, 
while  the  remaining  vertical  and  diagonal  web  members  are 
in  tension. 

As  shown  by  the  diagram,  the  total  length  of  the  trusses 
between  the  centers  of  end  pins  is  295.5  feet.  The  following 
statement  gives  all  the  dimensional  data  required  in  the  suc- 
ceeding computations : 

Center  depth  at  CP  =  42  feet. 
Depth  "     Ji  =  32     " 
"       -     P._  =  3of  - 
"       "     /^  -  29^  " 
"    Tio  =  28     " 
Length  of  center  panel  =  20  feet  6  inches. 
Panel  length  =  27.5  feet.  Total  length  =  295.5  feet. 

Point  of  intersection  of  upper  and  lower  chords  of  arms  is 
20  panels  from  the  free  extremity  of  each  arm. 

The  trusses  are  placed  16  feet  apart,  centers  transversely. 
The  trigonometric  quantities  required  are  : 


134 


SWING  BRIDGES. 


sec  for 

EP  = 

1-4 

t<       (( 

a  = 

1 .00 1 ,2 1 

<(      <( 

fi  = 

1.064 

((      (( 

T,= 

I-37I 

((      (( 

Ts  = 

1-343 

"      <' 

T-2   = 

1-319 

i(      <( 

cc  = 

1-397 

tan  for  £P  =  0.982 
"        a  =  0.048,5 

/3  =  0.364 
"  T,  =  0.938 
"  ^3  =  0.897 
"  7;  =  0.859 
*'       r^  =  0.976,2 


The  moving  load  to  be  used  consists  of  two  coupled  con- 
solidation locomotives,  identical  with  that  shown  in  Figure  i 
of  Article  77,  followed  by  a  uniform  train  load  of  3,000  pounds, 
or  1.5  tons,  per  lineal  foot,  the  moment  tabulation  for  which 
is  shown  on  page  41. 

The  fixed  load  will  be  as  follows  : 

Track 400  pounds  per  lineal  foot. 

Floor  beams  and  stringers  .      320       "  "        "         " 

Trusses 980       "  «        « 


Total 1,700       •' 

The  weight  of  trusses  will  be  taken  to  be  divided  equally 
between  the  upper  and  lower  panel  points,  but  the  track, 
floor  beams,  and  stringers  will  be  taken  wholly  at  the  latter. 
Hence  the  upper  and  lower  panel  fixed  loads  will  be  as  fol- 
lows : 


Lower  panel  load 

Upper      " 
Total    " 


1210 

2 
490 


X  27.5  =  16,650  lbs.  =    8.325  tons. 
X  27.5  =    6,750   "     =    3.375     " 


23,400 


=  11.700 


€t 


In  order  that  the  weight  of  the  locking  gear  at  the  extrem- 
ity of  each  arm  may  be  properly  provided  for,  the  weight  con- 
centrated at  the  free  extremity  of  each  truss  will  be  taken  at 
5  tons,  which  is  somewhat  more  than  one-half  of  each  of  the 
other  lower  panel  fixed  loads. 

The  panel  load  at  the  top  of  cP  will  be : 

2  2  ^  ^^ 


ENDS  SIMPLY   SUPPORTED.  135 

The  stresses  due  to  the  fixed  loads  given  above  will  first  be 
found.  Since  it  is  assumed  that  the  ends  of  the  arms  are  sim- 
ply supported  when  the  draw  is  closed,  the  stresses  arising 
from  the  fixed  or  own  weights  are  cantilever  stresses  identi- 
cal with  those  of  the  open  draw  ;  or,  in  other  words,  the  entire 
fixed  load  is  carried  at  the  two  points  of  support  over  the 
drum  on  the  pivot  pier.  As  the  chords  of  the  arms  are  not 
parallel,  all  web  stresses  will  be  found  by  taking  moments 
about  the  point  of  intersection  of  those  chords,  distant  20 
panels  from  the  extremity  of  either  arm,  as  stated  above. 
For  example,  in  order  to  find  the  stress  in  T^,  an  imaginary 
cut  or  section  is  to  be  taken  through  (^,  7^,  and  3  ;  it  is  then 
to  be  noted  that  the  lever  arm  of  T^  {i.e.,  the  normal  dropped 
from  the  intersection  of  b  and  3  on  T2,  produced)  is  22  x  27.5 
-i-  sec  for  Z3.  Now  the  lever  arm  of  the  five-ton  weight  at  the 
extremity  of  the  arm  is  20  panels  ;  that  of  the  panel  loads  act- 
ing along  TiQ,  21  panels;  and  that  of  the  loads  acting  along 
/*3,  22  panels. 

Hence  moments  give: 

.„,        5  X  20  +  11.7  X  (21  +  22)        _,  ,„ 

(7;)  =  ^— ^ — ^ ^  sec  7;  =  36.815  tons. 

The  chord  stresses  will  be  found  by  the  same  method  of 
moments,  the  center  of  moments  being  at  the  panel  point 
opposite  each  chord  panel — i.e.,  for  {b)  the  moment  center  is 
at  the  intersection  of  Psand  T^;  and  at  the  intersection  of  T^ 
andPj  for  (3)-  These  methods  are  precisely  those  outlined  in 
Article  17,  excepting  that  of  the  diagram,  which  is  not  used 
here. 

It  is  to  be  observed  that  all  the  stresses  can  be  found  by 
the  graphical  methods  given  in  the  preceding  pages,  but  the 
moment  method  only  will  be  used  in  this  example. 

It  will  be  assumed  that  the  counter  c^  sustains  no  fixed  load 
stress. 

Under  the  preceding  explanations,  the  desired  stresses 
will  be  as  follows  : 

(i)  =  —  5  tan  for  EP  =  —  4.91  tons. 


136  SWING  BRIDGES. 

/  X  5  X  55  +  II-7  ^  27.5 

(2)  =  — — ' —^  =  —  20.35  tons. 

^  ^  29.33 

i.\  5  X  82.$  +  11-7(55  +  27.5)_       ,,^,  ,^„. 

(3)  = -7^;^ =  -  44-93  tons. 

(4)  zzz  (5)  =  -  5  X  110+  11.7  (82.5  +  55  +  27.5) 

32 

=  —  77.52  tons. 

By  taking  moments  about  top  (or  foot)  of  CP: 

(6)  =-e=  -  5  X  137-5  +  117  (no  +  82.5  +  55  +  27.5) 

=  —  92.98  tons. 

{a)  =  —  (i)  sec  a  =  +    4.92  tons. 

{d)   =  —  (2)  sec  a  =  +  20.37     " 

(c)  =  —  (3)  seca  =  +  44.98     " 

(d)  =-         e  sec  /3  =  +  98.93     " 

(EP)  =  5  sec  for  EP  =  +  7  tons. 

(7-.o)  =  -(4f%^-:32|f^-)  =  -«•■— 

5  X  20  +  II. 7  X  21  +  3.375  X  22 


^P-^  =  - 


22 

=  —  19.09  tons. 


_        5  X  20  +  II. 7  (21  +  22)  +  3.375  X  23 

,      =  -  29.595  tons. 

Since  chord  d  cuts  the  lower  chord  3.2  panel  lengths  from 
foot  of  Zi,  or  0.8  panel  length  from  the  end  of  the  arm,  the 
center  of  moments  for  /*,  will  be  at  that  point,  while  the  lever 
arm  for  the  same  member  will  be  (4.2  panel  lengths  ~  sec  Pi). 
Hence: 


=  _  1 1-7  (Q-2  +  1.2  +  2.2  +  3.2)  -  5  X  0.8  ^^^  ^ 
4.2 
-  17.99  X  1. 319  =  —  23.73  tons. 


(^■)  =  4.^ 


ENDS  SIMPLY  SUPPORTED.  137 

The  stress  in  cP  is  simply  the  vertical  component  of  kU) 
added  to  the  fixed  load  at  the  top  of  cP.     Hence  : 

{cP)  —  —  {O  tan  (i  —  2.94  =  —  92.98   X  0.364  —  2.94 

=  —  36.785  tons. 

The  stresses  in  the  tension  members  T^,  T^,  and  T^  can 
now  readily  be  found  by  the  diagram,  Figure  2  of  Article  17, 
as  could  those  also  in  the  vertical  compression  web  members  ; 
but  the  moment  method  will  be  continued,  and  in  using  it  the 
center  of  moments  must  be  taken  at  the  point  of  intersection 
of  the  chords,  twenty  panel  lengths  to  the  left  of  the  foot  of 
EP.  The  lever  arm  of  7i  will  be  (21  panel  lengths  -^  sec  7^), 
and  similarly  for  the  otherinclined  tension  members.   Hence: 

,„,  5  X  20  +  II. 7  X  21  „ 

(T;)  =  +  • — sec  7;  =  +  22.57  tons. 

,^,             5x20+11.7(21+22)         _  .^ 

{Ts)=  + j^ ^  sec  Ts=  +  36.815  tons. 

.  ™-  5  X  20  +  1 1.7  (2 1  +  22  +  23)         „ 

(7;)  =  + ^-^ -^  sec  7;  =  +  50.02  tons. 

23 

(7i)  =  fixed  load  at  its  foot  —  +  8.325  tons. 

Since  there  is  no  fixed  load  stress  in  the  counter  Cj,  and  as  the 
members  LS'  and  cc  are  designed  simply  to  steady  the  bridge 
when  open,  and,  hence,  can  sustain  no  stress  that  can  be 
computed,  the  preceding  computations  complete  all  the  fixed 
load  truss  stresses. 

The  moving  load  stresses  are  next  to  be  found,  and  in 
finding  them  the  condition  of  "partial  continuity"  will  be 
assumed — /.r.,  it  will  be  assumed  that  no  shear  can  ever  pass 
from  one  arm  of  the  bridge  across  the  center  panel  to  the 
other.  The  web  members  in  the  center  panel  are  light  in 
section,  and  proportioned  only  to  hold  the  bridge  in  a  steady 
condition,  when  open,  against  any  tendency  to  vertical  vibra- 
tion by  the  wind,  or  irregular  or  uneven  operation  of  the  turn- 
ing machinery.     These  vibration  stresses  are  entirely  indeter- 


135  SWING  BRIDGES. 

minate  and  cannot  be  computed,  but  it  is  known,  from  actuai 
experience  with  drawbridges  of  great  length,  that  they  are 
small,  and  amply  resisted  by  rods  and  braces  quite  inadequate 
to  carry  more  than  a  very  small  portion  of  the  moving  load 
shear  that  would,  under  some  conditions  of  loading,  pass  the 
center  panel  if  the  trusses  were  perfectly  continuous  over  the 
drum.  Hence  such  a  construction  is  designed  as  will  not 
permit  the  existence  of  those  moving  load  shears  at  the  cen- 
ter, thus  leaving  only  the  incidental  stresses  of  turning  to  be 
cared  for. 

The  effect  of  this  arrangement  is  assumed  to  make  either 
arm  of  the  bridge  a  simple  truss  supported  at  each  end  for 
all  moving  loads  on  that  arm,  so  long  as  the  other  arm  carries 
no  moving  load. 

It  should  be  observed  that  the  span  4  (ordinarily  termed 
the  center  panel),  included  between  the  central  points  of  sup- 
port, sustains  no  load. 

If  the  moving  load  partially  covers  both  arms,  or  the  whole 
of  one  arm  and  a  part  of  the  other,  it  is  to  be  divided  into 
three  parts.  Two  of  these  parts,  together  with  the  reactions 
^1  and  y?4  due  to  them,  produce  equal  and  opposite  moments 
at  the  center.  For  these  two  parts,  consequently,  the  truss  is 
one  of  perfect  continuity,  with  the  main  diagonals  at  the 
center  omitted.  This  last  condition  is  admissible,  because  for 
these  two  parts  the  shear  at  the  center  will  be  zero. 

For  convenience  these  two  parts  will  be  called  "  balanced" 
while  the  third  part  will  be  called  "  unbalanced.'' 

For  the  unbalanced  part,  the  arm  or  span  in  which  it  is 
found  will  be  a  simple  truss  supported  at  each  end. 

If  the  panel  loads  are  uniform  in  amount,  balanced  loads 
will  be  symmetrically  placed  in  reference  to  the  center. 

If  the  panel  loads  are  not  uniform  in  amount,  the  balanced 
portions  would  be  determined  by  equating  M^,  to  M^,  with  the 
aid  of  Equations  (3)  and  (4)  of  Article  35. 

Coupling  these  statements  with  the  principles  deduced  in 
the  preceding  articles,  it  will  at  once  be  seen  that  the  greatest 
reaction  R-^2X  A,  Figure  i,  Plate  X,  will  exist  with  the  moving 
load  over  the  whole  of  the  left  arm,  for  any  moving  load  on 


ENDS  SIMPLY   SUPPORTED.  139 

the  right  arm  will  balance  a  part  of  that  on  the  left,  and  relieve, 
to  some  extent,  the  reaction  at  A. 

Although  the  moving  load  consists  of  the  various  wheel 
concentrations  followed  by  the  uniform  train,  as  shown  by  the 
following  diagram  ;  and  while  these  concentrations  can  read- 
ily be  used  for  the  moving  load  on  one  arm  only,  according 
to  the  principles  of  Articles  20  and  21,  their  use  in  the  same 
general  manner  for  balanced  moving  loads  would  lead  to 
excessive  complication.  Hence,  for  the  latter  loads,  a  fixed 
series  of  panel  weights,  resulting  from  one  position  of  the  dia- 
gram concentrations  given  above,  will  be  used  in  the  corre- 
sponding stage  of  the  computations.  By  this  device  the 
determination  of  the  stresses  will  be  much  simplified,  and  the 
results  will  be  essentially  accurate. 


gi 


£'        fil      £' 


■|^  ^r  ^i^ — "T— ^r — ^r-n 

9-'  jf  i'6'Y  i'&'^Y  -^^.  Y  1'10'>{>  .^8l^j^:iO^{»^^ 


Fig.  I. 

From  what  has  preceded,  it  is  clear  that  the  greatest  coun- 
ter stresses  are  to  be  found  by  carrying  the  moving  load  for- 
ward from  the  center  of  the  bridge  toward  the  end,  and  on 
one  arm  only,  at  the  same  time  considering  it  a  simple  non- 
continuous  span. 

The  greatest  main  web  stresses,  on  the  other  hand,  are  to 
be  determined  by  moving  the  system  of  balanced  panel 
weights,  already  described,  from  the  two  extremities  of  the 
bridge  toward  the  center,  as  will  presently  be  illustrated. 

It  is  to  be  remembered  that  verticals  in  compression  will 
sustain  their  greatest  stresses  at  the  same  time  with  the 
diagonal  tension  members  which  cut  their  upper  extremities, 
if  the  moving  load  traverses  the  lower  chord. 

As  one  arm  of  the  bridge  is  a  simple  truss  supported  at 


I40  SWING  BRIDGES. 

each  end,  for  the  unbalanced  loads  on  it,  it  is  evident  that  the 
greatest  compression  in  the  upper  chord  and  tension  in  the  lower 
will  exist,  near  the  ends,  for  the  moving'  load  over  the  whole  of 
one  arm. 

Since  moving  loads  on  both  arms  at  the  same  time  balance 
each  other,  it  results  that  the  greatest  tetision  in  the  upper  chord 
nnd  corApression  in  the  lower,  at  the  center,  ivill  exist  with  the 
moving  load  over  the  wJiole  of  both  arms. 

This  is  true  for  the  center  only.  For  other  panels  adjacent 
to  the  center,  //  zvill  be  necessary  to  take  single-balanced  pafiel 
moving  weights,  and  f^id  for  each,  all  panels  in  which  the  stress 
is  of  the  same  kind  as  that  caused  by  the  fixed  load  alone,  and 
the  amount  of  that  stress  in  those  panels. 

Having  obtained  the  results  for  each  pair  of  balanced 
weights,  they  are  to  be  combined  in  the  manner  already 
shown. 

The  greatest  compression  in  the  lower  chord  and  tension 
in  the  upper,  near  the  ends,  however,  will  exist  ivith  the  bridge 
open  or  closed  and  subjected  to  its  own  weight  only. 

From  these  results  the  greatest  chord  stresses  are  to  be 
found. 

The  maximum  moving  load  compression  in  EP  w'xW.  next 
be  found,  and  then  the  counter  stresses  and  that  in  Tig,  for 
all  of  which  the  arm  of  the  draw  will  be  treated  as  a  non-con- 
tinuous span,  with  the  moving  load  passing  over  it  from  the 
center  toward  the  end.  The  positions  of  the  moving  load  for 
the  greatest  value  of  these  stresses  will,  hence,  be  found  by  the 
principles  of  Articles  7,  20,  and  21.  Equation  (7)  of  Article 
7  shows  that  the  third  driver  of  the  front  locomotive  must  be 
placed  £t  the  foot  of  Tloin  order  to  give  the  greatest  shear  in 
EP,  or,  what  is  the  same  thing,  the  greatest  reaction  at  its 
foot.  Such  a  position  of  the  moving  load  will  make  the  shear 
in  question  45.16  tons.     Hence: 

{EP)  =  —  45.16  X  1.4  =  —  63.23  tons. 

In  order  to  determine  whether  the  counter  Ci  is  necessary, 
let  it  be  supposed  omitted,  and  the  resulting  greatest  moving 


ENDS  SIMPLY  SUPPORTED.  I4I 

load  compression  found  in  T^.  Equation  (^4)  of  Article  20 
shows  that  the  latter  will  exist  with  the  third  driver  of  the 
front  locomotive  at  the  foot  of  P^,  whence  will  result  a  reac- 
tion at  the  foot  of  ^5"/*  and  a  load  at  the  foot  of  Tio  of  32.1 
tons  and  5.72  tons,  respectively.  By  taking  moments  about 
the  point  of  intersection  of  the  chords,  and  using  the  same 
lever  arm  for  7^4  as  was  employed  for  the  fixed  load  stresses : 


=  —  34.08  tons. 

But  the  fixed  load  tension  in  T^  has  already  been  found  to 
be  22.57  tons,  which  is  much  smaller  than  the  34.08  tons  of 
moving  load  compression  just  found.  Hence,  as  Ti  can  really 
sustain  no  compression,  the  counter  c^  must  be  introduced,  and 
its  greatest  tension  found  with  the  same  position  of  moving 
load,  and,  also,  under  the  assumption  that  T^  does  not  exist. 
Remembering  that  the  lever  arm  of  c^  is  22  panel  lengths 
divided  by  the  secant  for  c,  : 

,    ,        \2.\    X  20  —  5.72  X  21 

(ci)  = ^-^ sec  c,,  =  +  23.72  X   1.4 

=  +  33.21  tons. 

By  a  precisely  similar  use  of  Equation  (4)  of  Article  20,  it 
will  be  found  that  the  greatest  moving  load  compression  will 
exist  in  7^3  with  the  first  driver  of  the  front  locomotive  at  the 
foot  of  Pg.  This  position  will  give  a  reaction  of  13.05  tons  at 
the  foot  of  £P,  and  an  advance  load  of  1 1.02  tons  at  the  foot 
of  P^.  Again  taking  moments  about  the  point  of  intersec- 
tion of  the  chords,  the  greatest  moving  load  compression  in 
Ts  will  be  found  to  be  : 

-^  13.05  X  20—  11.02  X  22         ^ 

(  io)  =  —  -^ — sec  Ts  =  —  14.45  tons. 

22 

But  the  fixed  load  tension  in  T3  has  already  been  found  to 
be  36.815  tons,  which  is  over  two  and  a  half  times  the  14.45 
tons  of  moving  load  compression.      Hence    no  compression 


142  SWING  BRIDGES. 

can  ever  exist  in  T^,  and  no  counter  will  be  needed  in  the 
same  panel  with  it,  nor,  indeed,  any  other  counter  than  c^. 

The  same  condition  of  loading  will  give  the  greatest  mov- 
ing load  tension  in  P^,  but  computations  will  show  that  it  is 
only  about  one-third  the  fixed  load  compression  (as  would 
be  inferred  from  the  results  for  T^  ;  hence  no  counterbrac- 
ing  of  that  member  is  required. 

These  operations  show  the  method  always  to  be  pursued 
in  determining  the  requisite  counters,  and,  also,  that  it  is 
essentially  identical  with  that  used  for  non-continuous  spans. 

As  T'lo  is  a  simple  hanger,  it  will  take  its  greatest  tension 
with  the  greatest  floor  beam  reaction  at  its  foot.  This  latter 
will  exist  with  the  same  position  of  moving  load  as  that 
which  gives  the  greatest  stress  in  EP — i.e.,  with  the  third 
driver  of  the  front  locomotive  at  the  foot  of  Tiq,  and  its  value 
is  25.57  tons.  Now  the  shear  existing  at  the  same  time  in 
EP  has  been  found  to  be  40.16  tons,  and  as  the  co-existing 
vertical  component  of  the  stress  in  a  is  between  three  and  four 
tons  only,  the  counter  Ci  must  act  in  tension  to  carry  sufficient 
shear  over  to  EP  to  make  up  its  shear  of  40.16  tons.  In 
other  words,  the  vertical  components  of  (^i)  and  (a)  added  to 
the  stress  in  Zlo  must  equal  40.16  tons.  Hence  the  stress  in 
Zio  is  simply  the  greatest  possible  load  at  its  foot. 

.-.  (r,o)  =  +  25.57  tons. 

The  member  d  sustains  no  stress,  under  the  condition  of 
partial  continuity,  with  the  moving  load  on  one  arm  only; 
hence  it  is  to  be  ignored  in  finding  the  stress  in  P^.  The 
greatest  compression  in  this  latter  member  will  therefore 
exist  with  precisely  the  same  position  of  moving  load  as  was 
used  for  EP,  but  changed  end  for  end  on  the  span,  so  as  to 
be  headed  toward  the  center  rather  than  toward  the  end  of 
the  arm. 

By  using  the  same  shear  as  for  EP,  therefore : 

(/*i)  =  —  45. t6  X  sec P■^  —  —  59.57  tons. 
The  tension  member  7\  is  a  simple  hanger,  with  the  maxi- 


ENDS   SIMPLY  SUPPORTED.  143 

mum  floor  beam  reaction  at  its  foot,  as  was  found  for  T^a- 
Hence: 

Ty^=  +  25.57  tons. 

The  moving  load  will  pass  on  the  arm  from  the  end 
toward  the  center,  for  the  remaining  web  members.  The 
position  of  moving  load  for  7*2,  in  accordance  with  Equation 
(4)  of  Article  20,  requires  the  third  driver  to  be  at  the  foot  of 
/g,  vvith  both  locomotives  and  tenders,  but  no  train  load,  on 
the  arm.  This  position  gives  a  reaction  of  32.1  tons  at  the 
foot  of  /'i,  and  an  advance  weight  of  5.72  tons  at  the  foot  of 
Zi.     Hence : 

_       32.1  X  25  —  5.72  X  24        ^  o    ^  . 

r,  =  ^— — ^ ^-^ ~  sec  T^  =  +  38.16  tons. 

2  23  ~  o 

In  the  same  manner  it  is  found  that  the  greatest  tension 
in  7^3  will  exist  with  the  second  driver  of  the  first  locomotive 
at  the  foot  of  P^,  and  with  the  corresponding  driver  of  the 
second  locomotive  on  the  arm  and  nine  inches  from  its  end. 
The  reaction  at  the  foot  of  /\  will  be  15.24  tons,  and  the 
weight  at  the  foot  of  /*2,  3.14  tons.  Hence,  by  moments,  as 
usual,  about  the  point  of  intersection  of  the  chords  : 

Since  Pc,  sustains  its  greatest  compression  under  the  same 
condition  of  moving  load  as  T^ : 

15.24  X  25 

Pi  = ^-^ ~  -  —  16.57  tons. 

I  23  ^' 

Again,  Equation  (4)  of  Article  20  shows  that  the  first  driver 
of  the  front  locomotive  must  be  at  the  foot  of  Tlo  in  order  to 
give  7^4  its  greatest  tension.  The  second  wheel  of  the  tender 
will  then  be  on  the  arm  and  one  foot  from  its  end.  Hence 
the  reaction  at  the  foot  of  P-^  will  be  4.62  tons,  and  the 
weight  at  the  foot  of  P-^,  i.i  tons,  and  : 

_       4.62  X  25  —  I.I  X  22        ^ 

7;  -  — —^ sec  7;  =  ^^  5.96  tons. 


144  SWING  BRIDGES. 

The  same  position  of  moving  load  gives  the  greatest  com- 
pression in  P^ ;  hence  : 

_,  4.62  X  25 

Pz=  — =  —  5.25  tons. 

^  22  ^    ^ 

These  results  complete  the  moving  load  web  stresses  for 
the  moving  load  on  one  arm  only,  leaving  the  chord  stresses 
yet  to  be  determined. 

The  lower  chord  panels  i  and  2  receive  their  greatest 
stresses  with  £P,  and  they  are  to  be  found  by  simply  multi- 
plying the  shear  in  that  member  by  the  tangent  for  £P. 
Hence : 

(i)  =  (2)  T=  45.16  X  0.982  =  +  44-345  tons. 

Similarly,  the  panels  4  and  5  sustain  their  greatest  stresses 
with  Pi.     Hence : 

(4)  =  (5)  =  45.16  X  0.859  =  +  38.79  tons. 

In  order  to  determine  the  greatest  stress  in  lower  chord 
panel  3,  recourse  must  be  had  to  Equation  (14)  of  Article  7. 
That  equation  shows  that,  with  the  moving  load  passing  on 
the  arm  from  right  to  left,  the  third  driver  of  the  second 
locomotive  must  be  at  the  foot  of  P^,  with  24  feet  of  the 
uniform  train  load  resting  on  the  arm  adjacent  to  the  center. 
This  position  gives  a  bending  moment  about  the  top  of 
P^  (the  center  of  moments  for  panel  3)  of  3,672,530  foot- 
pounds.     Hence  : 

3,672  5^0 
^^^  "^  ^  30.666     "  "^  121,070  lbs.  =  -f  60.535  tons. 

The  upper  chord  stresses  at  once  result  from  those  in  the 
lower,  but  it  is  first  to  be  observed  that  since  T^  is  a  tension 
member  only,  it  will  not  be  stressed  with  the  moving  load  on 
essentially  the  whole  of  the  arm,  since  the  counter  c^  will  then 
come  into  action.  The  upper  chord  stresses  in  a  and  d  will 
therefore  be  equal,  and  as  the  foot  of  P^  will  be  the  center 


ENDS  SIMPLY  SUPPORTED.  1 45 

of  moments  for  those  stresses,  the  position  of  the  moving 
load  for  their  greatest  value  will  be  precisely  the  same  as  for 
(3),  with  the  exception  that  the  direction  of  motion  is  to  be 
reversed,  thus  placing  the  third  driver  of  the  second  locomo- 
tive at  the  foot  of  Pg.     Hence  : 

5  672  Ej  ■^o 
{a)  =  {b)=  -  ~^~j-  seca=-  125,350  lbs.  =  -  62.675  tons. 

"y*  J  J  J 


Finally: 


[c)  =  —  (3)  sec  a  =  —  60.61  tons. 


These  computations  determine  all  the  greatest  moving 
load  stresses  in  one  arm  considered  as  a  simple  non-continuous 
span. 

In  finding  the  greatest  stresses  due  to  the  moving  load  on 
both  arms,  the  condition  of  partial  continuity  requires  the 
use  of  balanced  loads — t.e.,  loads  simultaneously  on  each  arm 
which,  if  the  trusses  were  continuous  over  the  center,  would 
produce  equal  and  opposite  center  bending  moments.  The 
most  obvious  and  simple  balanced  loads  are  those  of  equal 
magnitude  placed  at  symmetrical  panel  points  in  each  arm, 
and  such  balanced  loads  will  be  used  in  the  following  com- 
putations. 

As  has  already  been  observed,  the  use  of  the  locomotive 
concentrations  and  the  ordinary  moment  diagram  for  the 
greatest  stresses  in  this  case,  would  lead  to  excessive  compli- 
cation and  great  labor  without  any  corresponding  advantage. 
Essential  accuracy  can  be  attained  by  computing  a  system 
of  panel  concentrations  or  weights  from  a  position  of  the 
moving  load  which  will  give  results  differing  in  no  sensible 
degree  from  those  determined  by  the  most  refined  calcula- 
tions. The  desired  position  is  largely  a  matter  of  judgment, 
but  a  heavy  concentration  should  be  found  at  the  panel 
point  covered  by  the  head  of  the  train.  In  the  present 
instance  the  third  driving  wheel  of  the  front  locomotive  will 
be  placed  over  a  panel  point  which  will  be  called  panel  i. 
A  small  concentration  will,  of  course,  exist  at  the  panel  point 
10 


14^  SWING  BRIDGES. 

in  front  of  panel  i  ;  this  will  be  called  the  "  advance  load.** 
The  desired  concentrations  will  then  be  as  follows: 

Advance  load,  5.70  tons. 

At  panel  i,  25.55      " 

"       -       2,  17.75       " 

"       "      3,  25.55      " 

•'      "      4,  19-35      " 

The  advance  load  is  sometimes  neglected  in  finding  the 
web  stresses,  as  the  resulting  computations  are  somewhat 
simplified,  and  the  small  error  committed  is  on  the  side  of 
safety. 

All  reactions  in  this  case  must  be  found  by  the  equations 
of  the  preceding  Article  for  perfect  continuity,  as  all  loads 
are  to  be  balanced.  If  a  panel  load,  /*,  rests  on  the  left-hand 
arm,  or  span,  /j,  the  reaction  R^  will  be  by  Equation  (5«)  of 
Article  35  : 

7?.  =  /'(.-»)|'-('  +  ")"- — - — r\  ■  (I). 

(  4(^+.)-,-^f 

And  by  Equation  {6d)  of  Article  35,  the  reaction  R^  will 
be: 

R,  =  P{\-7i^)n      .    J .^  .     .     .     (2). 

In  this  case,  c  =  137.5  -^  20.5  =  6.7073. 

If  the  panel  points  be  numbered  from  i  to  4  from  the  end 
of  the  arm  to  the  center,  panel  i  being  at  the  foot  of  Zlo,  and 
panel  4  at  the  foot  of  Zi ;  and  if  P  be  the  general  value  of  a 
panel  load,  n  will  have  the  values  0.2,  0.4,  0.6,  and  0.8  in  the 
formulae  for  Ri  and  R^,  and  the  expressions  for  those  reactions 
will  be  : 

Panel  load  Pat  panel  i,  R^  =  0.716102  P;  R^  =  0.005443  P. 

"      2,  i?i  =0.453178/';  7?4  =  0.009526/'. 

"         "       "         "      3,  i?i  =  0.232204  P;  /?4  =  0.010886  P. 

"         "       "         "      4,  7?i  =  0.074153  P;  i^4  =  0.008165  P 


ENDS   SIMPLY  SUPPORTED.  1 47 

Since  balanced  loads  only  are  to  be  employed  in  this  case, 
It  will  be  necessary  to  determine  the  reactions  R^  and  Ri  with 
a  panel  concentration  on  one  arm  balanced  by  an  exactly 
equal  concentration  symmetrically  placed  on  the  other.  The 
reactions  for  each  such  concentration  will  be  those  given 
above,  but  R^  and  Ri  will  be  interchanged  with  the  exchange 
of  one  arm  for  the  other.  When  the  two  concentrations  are 
simultaneously  placed  on  the  two  arms — i.e.,  when  they  are 
balanced — the  reaction  at  each  end  of  the  bridge  will  equal 
that  at  the  other,  and  will  be  represented  by  the  sum  of  R^ 
and  -^4,  as  given  above  for  a  single  panel  load.  If  R  is  that 
reaction  : 

For  panel  i,  R  =  Ry  +  R^  =  0.7215  P. 

"     2,  R  =  R,  +  R^  =  0.4627  P. 

"     3,  R=  R^  +  R,  =  0.2431  P. 

"     4,  R  =  R^  +  R^  =  0.0823  P. 

The  preceding  numerical  work  can  be  thoroughly  checked 
by  finding  M2  for  both  arms  fully  loaded,  from  Equation  (4) 
of  the  preceding  Article  (having  previously  found  J/g),  and 
using  it  in  Equation  (5)  of  the  same  Article.  The  result- 
ing value  of  Ri  =  Ri  should  equal  the  sum  of  the  preceding 
values  of  R,  because  both  results  are  for  both  arms  fully 
loaded.  The  sum  of  the  four  values  of  R  is  1.5096  P,  while 
Equation  (5)  of  Article  35  gives  R^  =  R  =  1.5097/*.  Hence 
the  check  is  satisfactory.  In  this  verification  all  panel  loads 
P  are  assumed  equal,  but  this  does  not  in  any  way  affect  the 
numerical  work  which  it  was  desired  to  verify. 

As  the  reactions  for  each  of  the  panel  loads,  and  the  panel 
loads  themselves,  are  now  known,  all  stresses  can  readily  be 
found.  In  all  the  cases  except  those  of  simple  hangers,  or 
where  a  trigonometric  multiplier  only  is  needed,  the  method 
of  moments  will  be  employed,  precisely  as  with  the  fixed  load 
and  with  the  moving  load  on  one  arm  only. 

The  greatest  stress  in  EP  will  evidently  exist  with  the  great- 
est reaction  at  its  foot — i.e.,  with  the  moving  load  over  the 
whole  of  both  arms.     Hence  the  reaction  is  : 


148  SWING  BRIDGES. 

07215    X  25.55  =   18.435 

0.4627  X  17.75  =     8.210 

0.2431  X  25.55  =    6.210 

0.0823  X  19.35  =     1.595 

.-.     R   =  34.450  tons. 
Hence:      {EP)  =  —  34.45  x  sec  EP  —  —  48.23  tons. 

The  same  condition  of  loading  gives  the  greatest  stress  in 
lower  chord  panels  i  and  2  ;  therefore  : 

(i)  =  (2)  =  34.45  x  tan  EP^  +  33.83  tons. 

The  hanger  T^q  will  be  treated  precisely  as  in  the  case  of 
moving  load  on  one  arm  only,  but  with  the  balanced  concen- 
trations.    Hence : 

(7;o)=  +  25.55  tons. 

The  position  of  loading  for  the  greatest  stress  in  c-^  is  with 
the  advance  weight  5.7  tons  at  the  foot  of  /"lo,  as  the  resulting 
reaction  is  then  greater  in  comparison  with  the  load  between 
the  end  of  the  arm  and  foot  of  c^  than  with  any  other  position. 
If  the  advance  load  were  very  large  it  would  have  to  be 
placed  at  the  foot  of  c^  for  the  greatest  stress  in  that  mem- 
ber.    The  reaction  then  becomes : 

0.7215  X  5.7    =  4.112  tons. 

0.4627  X  25.55  —  J 182 

0.2431  X  17.75  ^  4.315     " 

0.0823  X  25.55  =     2.103    " 

R  =  22.350    " 

Remembering  that   T^  is  now  to  be  neglected  : 

,   ,       22.35  X  20  -  5.7  X  21  ,         _     ^ 

{c^  =  — -^^ ■^-^ sec  c^—+  20.83  tons. 

Since  the  stress  in  c^  is  much  less  than  was  found  with  the 
moving  load  on  one  arm,  as  was  to  be  anticipated,  it  is  clear 
that  no  other  counter  stresses  need  be  found.  Indeed,  it  is 
evident  from  the  general   conditions  of  the  two  cases  that 


ENDS  SIMPLY  SUPPORTED.  1 49 

the  greatest  counter  web  stresses  must  be  found  with  the 
moving  load  on  one  arm  only,  as  has  already  been  observed. 

The  conditions  of  loading  for  the  greatest  stresses  in  the 
main  web  tension  members  7^,  To,,  and  T^  will  be  found  by 
bringing  the  moving  load  on  the  bridge  from  the  end  of  the  arm 
toward  the  center,  every  panel  concentration  at  the  same  time 
being  balanced.  In  the  present  case,  the  advance  load  of  5.7 
tons  is  to  be  placed  one  panel  in  front  of  the  foot  of  the  mem- 
ber in  question,  as  it  is  small  in  comparison  with  that  which 
follows  it.  This  advance  load  might  be  so  large  as  to  require  it 
to  be  placed  at  the  foot  of  the  member  whose  stress  is  sought. 
The  choice  between  these  two  positions  can  only  be  deter- 
mined by  trial.  One  of  these  positions  of  the  moving  load 
will  give  the  greatest  stress  desired,  for  the  reason  that  the 
end  reaction  will  then  be  the  least  in  comparison  with  the 
moving  load  between  it  and  the  member  considered,  thus 
insuring  the  greatest  possible  shear  in  the  latter. 

It  is  to  be  observed  that  the  design  of  the  truss  is  such 
that  the  inclined  web  members  are  subject  to  pure  tension, 
and  the  vertical  posts  P^  and  P^  to  pure  compression, 
necessitating  the  introduction  of  the  counter  c-^.  If  it  is 
desired  to  omit  the  latter  member,  making  a  rather  more 
excellent  design  from  a  purely  engineering  point  of  view,  T^, 
and  perhaps  other  adjacent  inclined  web  members,  will  be 
subjected  to  compression,  with  possibly  some  of  the  vertical 
posts  in  tension,  and  all  such  members  must  be  counterbraced. 
The  requisite  computations  for  this  case  will  be  considered  at 
the  end  of  this  Article. 

In  order  to  find  the  greatest  stress  in  T^,  the  advance  load 
of  5.7  tons  is  to  be  placed  at  the  foot  of  P^\  hence  the 
reaction  R  will  be  : 

0.7215  X  25.55  =  18.435 
0.4627  X     5.7    =    2.637 

K.       =  21.072  tons. 


,_.       25.55  X  21  -  21.072  X  20  ,.. 

.-.  (T;)  =  -^^^ ^^^ '—       se<  /4  =  +  7-5 1  tons. 


150  SWING  BRIDGES. 

The  advance  load  must  be  at  the  foot  of  P^  for  the  greatest 
stress  in  /"g,  and  the  resulting  reaction  will  be  : 

0.7215  X  17.75  =  12.808 
0.4627  X  25.55  =  11.820 
0.2431  X     5.7    =    1.385 

R      =  26.013  tons. 

25.55  X  22  +  1775   X  21  -  26.013  X  20    _  ^ 

—  +  25.315  tons. 

For  the  greatest  stress  in  T^,  the  advance  load  must  be  at 
the  foot  of  Tx ;  whence  : 

0.7215  X  25.55  ^  18.435 

0.4627  X  17.75  —    8.212 

0.2431  X  25.55  =    6.213 

0.0823  X      5.7     =     0.470 

^      =  33-330  tons. 
25 . 5 5  X  2 3  +  17.75  X  22  +  25.55  X  21  -  33.33  X  20 

sec  T.2  =  +  48.64  tons. 


•••  (^^^  23 


In  illustration  of  the  effect  on  the  stress  in  T,  of  placing 
the  advance  load  at  its  foot,  rather  than  at  the  foot  of  7",, 
the  following  would  be  the  value  of  that  stress,  remembering 
that  the  reaction  R  would  be  26.013  tons,  as  found  for  T^: 

5.7  X  23  +  25.55  X  22  +   17.75  X  21  -  26.013   X  20 
(A)  -  23 

sec  T.i=  +  31.3  tons, 

which  is  seen  to  be  about  two-thirds  of  the  greatest  stress. 
The  result  would  have  been  very  different,  however,  if  the 
5.7  tons  were  displaced  by  a  large  concentration. 


ENDS  SIMPLY  SUPPORTED.  151 

As  Tx  is  a  simple  hanger,  it  receives  only  the  greatest  load 
at  its  foot : 

(Ti)  =  +  25.57  tons. 

The  greatest  reaction  at  the  foot  of  /\  will  exist  when  the 
advance  load  of  5.7  tons  is  placed  at  that  point,  and  its  value 
will  be : 

0.7215  X  19.35  =  13.962 

0.4627  X  25.55  =  11.820 

0.2431  X  17.75  =    4.315 

0.0825  X  25.55  =    2.103 

R      —  32.200  tons. 

Since  the  upper  chord  d  cuts  the  lower  chord  panel  i  at 
a  point  0.8  panel  length  from  the  foot  of  EP,  that  point  of 
intersection  must  be  taken  as  the  center  of  moments  for  /^j ; 
hence  : 

(/'O  = 

32.2x0.8+19.35  X  0.2 +  25.5 5  X  1.2  + 17.75  X  2.2  +  25.55  x3'2 

4,2 

sec  P^  =  —  56,875  tons. 

The  posts  P2  and  P^  take  their  greatest  stresses  with  the 
web  members  T^g  and  T^  respectively  ;  hence  the  reactions 
for  the  latter  members  are  to  be  used  for  the  former.  There- 
fore : 

iD\  25.55x22  +  17.75x21—26.013x20  o 

{Pi)= '^-^ —^-^ =  —  18.025  tons. 

Also: 

fr,\  25.55  X  21  —  21.072  X  20 

(^3)  =  -  -^^ Yi =  ~  5.235  tons. 

These  complete  the  web  stresses,  and  leave  only  those  for 
the  chords  [except  (i)  and  (2)  already  found]  to  be  deter- 
mined. As  the  counter  c^  comes  into  action  with  the  moving 
load  over  the  whole  of  the  bridge,  the  two  upper  chord  panels 


152  SWING  BRIDGES. 

a  and  b  will  sustain  the  same  stress,  and  panel  point  2  at 
the  foot  of  P2,  will  be  the  center  of  moments  for  their  stress. 
Now  if  each  panel  load  be  called  unity,  the  reactions  Rx  ot 
the  loads  at  panel  points  i,  2,  3,  and  4  will  be  0.7215, 
0.4627,  0.2431,  and  0.0823  respectively  ;  and  if  each  of  these 
reactions  multiplied  by  its  distance  from  the  center  of 
moments  be  greater  than  the  product  of  the  panel  load 
that  produces  it  multiplied  by  its  lever  arm,  the  effect  of 
placing  that  load  on  the  span  will  be  the  production  of  a 
stress  of  the  same  kind  as  that  due  to  the  reaction  alone. 
If  the  moments  of  these  unit  reactions  and  that  of  the  unit 
load  at  panel  point  i  be  taken  about  panel  point  2,  there  will 
result  : 

For  panel  load  I,  (0.7215  x  2  =  1-443)  —  i  =  -I-  0.4430 

"          "         "       2,  0.4627  X  2                            .=  +  0.9254 

"            "          "        3,  0.2431  X   2                                 =    +  0.4862 

"            "          "        4,  0.0823  X    2                                   =   +  0.1646 

Since  all  these  unit  moments  are  positive,  all  loads  will 
produce  compression  in  a  and  b.  It  is  clear,  also,  that  the 
heaviest  load  must  be  placed  at  the  foot  of  P^\  this  will 
locate  the  two  heaviest  concentrations  at  the  feet  of /s  and  Zi, 
with  the  head  of  the  train  toward  P^.  The  actual  moments 
of  the  panel  loads  about  the  foot  of  P^  can  now  readily  be 
found  by  multiplying  those  loads  by  the  unit  moments  given 
above,  and  then  by  multiplying  the  sum  of  the  results  by  the 
panel  length,  as  follows  : 

19.35  X  0.4430  =    8.572 

25.55  X  0.9254=  23.644 

17.75  X  0.4862  =    8.630 

25.55   X  0.1646=    4.206 

Total  moment     .     .     .     45.052  x  27.5. 
Hence  the  stress  in  a  and  b  will  be  : 

(^)  =  (^)=-45-Q52x27,5 
29.333 
Y.  sec  a  =^  —  45.052  X  tan  T^  a  sec  a  —  —  42.31  tons. 


ENDS  SIMPLY  SUPPORTED.  153 

In  order  to  find  the  greatest  stress  in  lower  panel  3, 
Tnoments  must  be  taken  about  the  top  of  Pj.  By  taking 
unit  moments,  as  before,  about  that  center: 

For  panel  point  i,  (0.7215  x  3  =  2.1645)  —  2  =:  +  0.1645 

"       2,  (0.4627  X  3  =  1. 3881)  —  I  =  +  0.3881 

"       3,  0.2431   X  3                              =  +  0.7293 

*'         "          "       4,  0.0823  X  3                              =  +  0.2469 

Since  all  these  results  are  positive,  all  loads  on  the  arm  will 
produce  tension  in  lower  panel  3,  or,  what  is  the  same,  com- 
pression in  upper  panel  c.  For  the  greatest  stress  in  panel 
3,  one  of  the  heaviest  concentrations  must  be  placed  at  the 
foot  of  P^_,  so  that  the  two  heaviest  concentrations  will  be 
located  at  the  feet  of  P^  and  r,o  with  the  head  of  the  train 
toward  EP.     Hence  : 

25.55  X  0.1645  =.    4.203 

17.75  X  0.3881  =    6.889 

25.55  X  0.7293  =  18.633 

19.35  X  0.2469  =    4.778 

Total  moment     .     .     .     34.503  x  27.5. 

(3)  =  34-503  X  ^^   =  34.503  X  tan  7;  =  +  30.95  tons. 
Also  :  ic)  =  —  (3)  X  sec  a  —  —  30.985  tons. 

Unit  moments  about  the  top  of  P^  for  the  stress  in  lower 
panels  4  and  5  will  result  as  follows  : 

For  panel  point  i,  (0.7215  x  4  =  2,8860)  —  3  =  —  0.1140 

"         "         "       2,  (0.4627  X  4  =  1.8508)  —  2  =  —  0.1492 

"         "         "       3,  (0.2431  X  4  =  0.9724)  —  I  =  —  0.0276 

"         "         "      4,  0.0823  X  4                              =  +  0.3292 

These  results  show  that  the  panel  load  at  the  foot  of  7i  is 
the  only  one  which  produces  tension  in  panels  4  and  5  ; 
while  it  is  evident  that  one  of  the  heaviest  concentrations 
should  be  placed   at  the  foot  of  7i.     Hence,  by  placing  the 


154  SWING  BRIDGES. 

advance  load   of  5.7  tons  at  the   foot  of  T^,  and   the  heavy 
load  of  25.55  tons  at  the  foot  of  Zl : 

For  panel  3,       57     x  (— 0.0276) —— 0.1575 
"         "      4,     25.55  X         0.3292   =  +  8.41 1 


Total  moment     .     .     .     8.2535  ^  27.5. 


Hence 


(4)  =  (5)  =  8.2535  X  ^  =  8.2535  X  tan  /'i  =  +  7.09  tons. 
32 


In  order  to  find  the  greatest  compression  in  panels  4  and  5,. 
the  advance  load  of  5.7  tons  is  to  be  placed  at  the  foot  of 
7*2  with  the  head  of  the  train  toward  /\.      Hence: 

For  panel  i,  '17.75  x  (—0.1140)  =  —  2.024 
"  2,  25.55  X  {- 0.1492)  =-  3.812 
"       3,        5.7     X  (- 0.0276)=  -0.157 

Total  moment     .     .     .     5-993  x  27.5. 
Hence:       (4)  =  (5)=  —  5-993  x  tan  P^  =  —  5.15  tons. 

The  advance  load  might  be  so  small  that  it  would  be  neces- 
sary to  place  it  at  the  foot  of  7i  for  the  greatest  compression 
in  4  and  5. 

In  order  to  determine  the  greatest  stresses  in  panels  e  and 
6,  let  unit  moments  be  taken  about  the  foot  of  P-^ : 

For  panel  i,  (0.7215  x  5  =  3.6075)  —  4  =  —  0.3925 

"      2,  (0.4627  X  5  :=  2.3135)  -  3  =  -  0.6865 

"      3,  (0.2431   X  5  =  1.2 155)- 2  =  -0.7845 

''      4,  (0.0823  X  5  =  0.41 1 5)  -  I  =  -  0.5885 

These  results  show  that  all  loads  produce  tension  in  d  and 
e  and  compression  in  6.  Hence,  by  placing  the  two  heaviest 
concentrations  at  the  feet  of  T^  and  P^  with  the  head  of  the 
train  toward  P^ : 


ENDS   SIMPLY  SUPPORTED. 


155 


-  19-35  X  0.3925 

-  25.55  X  0.6865 

-  17-75  X  0.7845 
-25.55  X  0.5885 


/.  (6)  =-{e)  =  ~  54.096  X  ^  = 

42 


7-595 
17.540 

13-925 
15-036 

^•096 
-  35.42  tons. 


(^=  {e)  sec  ft  =  +35.41  X  1.064  =  +  37.685  tons. 

{cP)  —  —  {e)  tan  ft  —  —  35.41  x  0.364  =  —  12.895  tons. 

These  complete  all  the  moving  load  stresses,  and,  with 
those  due  to  the  fixed  load,  they  enable  all  the  resultant 
maxima  stresses  to  be  at  once  written.  The  following  tabu- 
lation shows  the  results  of  all  the  computations  : 

TABLE    I. 


Fixed  Load. 

Moving  Load  on 

One  Arm. 

Both  Arms. 

T, 
T^ 

T, 
T, 

+    8.325 
+  50.02 

+  36-815 

+  22.57 
-    8.14 

+  25-57 

+  38.16 

\  +18.85 

/  -14-45 

)  +    5-96 

{  -34-oS 

+  25-57 

+  33-21 

—  63.23 

-59-57 
-16.57 

-  5-25 

+  25-57 
+  48 . 64 

+  25-315 

+    7-51 

+  25-55 
+  20.830 
-48.23 
-56.875 

—  18.025 

-  5-235 
-12.895 

-42.31 

-42-31 

-30.985 

+  37685 

+  35-42 

+  33-83 

+  33-83 

+  30.95 

i  +    7-09 

\-    5-15 

i  +    7-09 

I-    5-15 

-35-41 

EP 

Px 

Pi 
P3 
cP 

+    7.00 
-23-73 
-29-595 
-19.09 

-36.785 
+   4-92 
+  20.37 
+  44.98 
+  98.93 
+  92.98 

-  4-91 
-20.35 

-44-93 

-77-52 

-77-52 
-92.98 

a 
b 
c 
d 

-62.675 

—  62.675 

—  60.61 

I 
2 

3 
4 

5 
6 

+  44-345 
+44-345 
+  60.535 

+  38-79 
+  38-79 

156 


SWING  BRIDGES. 


The  stresses  in  Table  I  show  that  the  tension  member  T^m 
must  be  counterbraced  so  as  to  sustain  the  tension  of  19.48 
tons  due  to  the  moving  load  on  one  arm  and  the  fixed  load 
compression  of  8.14  tons.  T^,  is  shown  under  a  compression 
of  14.45  tons,  due  to  moving  load  on  one  arm,  but  the  fixed 
load  tension  of  36.815  tons  is  more  than  two  and  one-half 
times  as  great ;  hence  no  counterbracing  would  be  necessary 
in  any  case.  Although  34.08  tons  compression  is  shown 
opposite  7*4,  it  belongs  to  the  case  in  which  c^  does  not  exist, 
and  which  will  be  treated  at  the  end  of  this  Article  ;  it  there- 
fore needs  no  further  consideration  here.  The  entire  lower 
chord  must  be  counterbraced,  as  must  also  the  upper  chord 
panels  a,  b^  and  c.  These  latter  members  are  always  built 
with  sections  for  compression.  Table  II  shows  the  final 
resultant  stresses  desired. 

TABLE    II. 


Member. 

Tension. 

Compression. 

Member. 

Tension. 

Compression. 

+    33-895 
+    98   66 
+    62.13 
+    30.08 
+    17-43 
+   33-21 
+     7.00 

a 

b 

!/ 

e 

\             I 

i                2 

3 
4 

5 
6 

+      4.92 
+    20.37 
+    44-98 
+ 136.615 
+ 128.39 
+   39-435 
+   23.995 
+    15.605 

-  57-755 

-  42.305 

-  15-63 

Ti 

Ti 

/"lo 

—      8.14 

EP 

-  4-91 

-  20.35 

-  44-93 

-  82.67 

-  82.67 
-128.39 

-  56.23 

-  83.30 

-  47-62 

-  24.34 

-  49-68 

Pi 

Pi 

cP 

It  is  noticeable  in  Table  I  that  P^  receives  greater  stress 
with  moving  load  on  one  arm  only,  than  on  both  arms.  This 
is  in  consequence  of  the  inclination  of  the  upper  chord  panel 
d,  by  which  it  takes  a  considerable  shear  that  would  other- 
wise exist  in  /'j,  and  which  would  correspondingly  augment  its 
stress.  The  same  Table  also  shows  that  the  greatest  stresses 
exist  in  the  various  portions  of  the  arm  under  the  conditions 
of  loading  indicated  in  a  former  portion  of  this  Article: 

The  preceding  constitutes  all  that  pertains  to  the  computa- 


ENDS  SIMPLY   SUPPORTED.  157 

tion  of  stresses  in  the  trusses,  but  it  is  still  necessary  to  com- 
pute the  reactions  R^  and  R^,  at  the  extremity  of  the  arm 
and  over  the  center  pier  respectively.  The  supports  at  the 
latching  and  locking  points  must  be  designed  to  resist  the 
reaction  ^,,  and  the  drum  must  be  designed  to  support 
the  reaction  R.,.  R^  will  manifestly  have  its  greatest  value 
with  the  moving  load  on  one  arm  only.  The  determination 
by  trial  of  the  maximum  reaction  at  the  foot  of  EP,  including 
the  proportionate  part  of  those  wheel  concentrations  in  the 
panel  adjacent,  is  somewhat  laborious,  and  as  each  locomo- 
tive and  tender  weight  divided  by  its  total  length  averages 
3,152  pounds,  or  1.576  tons,  per  lineal  foot,  it  will  be  essen- 
tially accurate  and  much  simpler  to  assume  that  the  total 
maximum  reaction  to  be  supported  at  the  locking  and  latch- 
ing point  at  the  foot  of  EP'\s\ 

^^  ^  ^37-5  X  0.775  ^  6  ^  5^  ,8  ,^„3 

The  concentration  of  6  tons  is  added  to  provide  for  the 
driving-wheel  concentration  that  must  be  supported  at  the 
instant  of  entering  on  or  leaving  the  arm  while  the  whole  of 
the  latter  is  otherwise  loaded. 

The  reaction  for  both  arms  covered  with  moving  load  is 
best  determined  in  precisely  the  same  manner.  Each  panel 
load  will  be  27.5  x  0.775  =  21.312  tons.  Now  the  reaction 
i?,  for  four  such  panel  loads  will  be  : 

(.7215  +  .4627  +  .2431  +  .0823)  X  21.312  =  1.5096  X  21.312. 

.'.     R^  =  32.173  tons. 
Hence  : 

7?,'  —  4  X  21.312  —  32.173  +  — ~  X  21.312  -\-  6  —  77.6^^  tons. 

The  panel  load  added  is  for  the  half  panel  adjacent  to  the 
center  and  the  central  panel,  while  the  6  tons  is  for  the  driv- 
ing-wheel concentration,  as  just  explained  in  connection 
with  R..     To  R'  must  be  added  the  total  fixed  load  carried 


158  SWING  £  JUDGES. 

to  the  center.     From  the  data  given  at  the  beginning  of  this 
Article,  that  total  fixed  load  will  be: 

24  X  0.85       , 
4  X  11.7  +  5  +  -^ ^  =  62.0  tons. 

Hence  the  total  reaction  R^  to  be  supported  at  the  drum 
will  be  : 

R^  =  77 '^74  +  62  =  139.674  tons. 

As  the  ends  of  the  span  are  simply  supported,  no  fixed 
load  reaction  is  to  be  added  to  the  value  of  R^  as  determined 
by  the  moving  load  only. 

The  omission  of  counters. 

It  has  already  been  remarked  that  the  best  design,  from  a 
purely  engineering  point  of  view,  is  secured  by  omitting  the 
counters  and  counterbracing  the  main  web  members.  In  the 
present  instance  the  counter  <:-  would  be  omitted  and  the 
main  members  7^  and  P^  counterbraced.  The  member  T^^ 
will  also  receive  much  greater  tension  than  in  the  preceding 
case. 

In  considering  these  counter  stresses,  the  arm  of  the 
bridge  is  to  be  treated  as  a  simple  non-continuous  span,  as 
was  done  in  the  earlier  part  of  the  Article,  with  the  moving 
load  passing  from  the  center  toward  the  end.  Equation  (4) 
of  Article  20  shows  that  the  third  driver  of  the  front  locomo- 
tive should  be  at  the  foot  of  7"]o  in  order  to  give  that  mem- 
ber its  greatest  tension.  The  first  parenthesis  of  the  second 
member  of  that  equation  disappears  in  this  case  because  in 
is  zero.  The  resulting  reaction  at  the  foot  of  EP  is  50.86 
tons,  of  which  5.70  tons  is  due  to  the  wheel  concentrations 
between  the  foot  of  EP  and  the  foot  of  Zio-  Hence,  by 
moments  about  the  chord  intersection  : 

^        (50.86  —  5.70)  20 

Tio  =  — ±-L-L —  =  +  43.01  tons. 

The  moving  load  compression  in  T^  has  already  been  found 
to  be  34.08  tons,  in  deciding  upon  the  necessity  for  c^  in  the 


ENDS   SIMPLY  SUPPORTED.  159 

preceding  case.     The  same  position  of  the  moving  load  (the 

third  driver  at  the  foot  of  P.^  will  give  the  greatest  tension 

in  /g,  and  using  the  data  already  employed  for  Ti,  there  will 

result : 

„       32.1  X  20  —  5.72  X  21 
P,  =  ^ ^^ =  +  23.73  tons. 

Again,  it  was  found  that  the  greatest  moving  load  compres- 
sion in  Tg  would  be  14.45  tons,  while  the  fixed  load  tension  is 
36.815  tons  ;  hence  T^  can  never  suffer  compression.  The  re- 
sults for  7^3  show  also  that  the  fixed  load  compression  in  P^  will 
always  largely  overbalance  any  possible  moving  load  tension. 
The  only  counterbracing  needed,  therefore,  will  be  that  for 
the  members  Ti  and  P3.  These  computations,  in  combina- 
tion with  the  results  given  in  Table  I,  show  that  Zio,  Ti,  and 
Pg  will  sustain  the  following  resultant  stresses  : 

7^10 ;    +  34.87  tons.     —    8.14  tons. 
Ti ;    +  30.08     "         —  1 1.5 1     " 
P, ;    +    4.64     "         -  24.34     " 

All  the  other  stresses  remain  unchanged. 

Art.  37. — Ends  Simply  Supported — Two  Points   of  Support  at  Center — 
Complete  Continuity — Example. 

The  case  of  complete  continuity  to  be  considered  in  this 
Article  involves  the  existence  of  web  members  in  the  panel 
over  the  drum  (?>.,  the  span  4  of  Article  35),  designed  by 
actual  computation  to  take  the  shear  which  may  pass  the 
center  when  the  structure  carries  unbalanced  loads. 

It  has  been  shown,  in  Article  35,  that  when  the  condition 
of  continuity  is  fulfilled,  which  is  the  only  condition  there 
contemplated,  either  of  the  reactions  R^  or  .^3  may  be  nega- 
tive, i.e.,  downward,  while  both  the  reactions  i?,  and  P^  are 
always  positive,  or  upward.  It  will  be  found  that  provision 
of  the  nature  of  heavy  anchorage  must  be  made  in  order  to 
meet  the  requirements  of  P2  and  ^3  with  some  conditions  of 
loading  of  draw  spans.  This  indicates  that  the  web  members 
in  the  center  panel  will  be  very  heavy,  and  such  will  be  found 
to  be  the  case. 


l6o  SWING  BRIDGES. 

The  moving  load  stresses  in  all  cases  will  be  determined 
by  means  of  the  formulae  of  Article  35.  The  truss  to  be  con- 
sidered is  the  same  as  treated  in  the  preceding  Article,  and 
is  shown  by  Figure  i  of  Plate  X.  The  data,  reproduced 
from  Article  36,  are  as  follows : 

Length  between  centers  of  end  pins  .     .     295.5  feet. 

Panel  length  in  arms 27,5     " 

Length  of  center  panel  or  span     ...       20        " 

Center  depth  at  CP  —  42     feet.     Dep*-h  at  Ti  =  32     feet. 
Depth  at  Pg  =  30|     "  "        "    P^  =  2g\     " 

"        "  Zio  —  28       "        Truss  centers  16  ft.  apart. 

Weight  of  rails,  ties,  and  guards,  with  bolts  and  connec- 
tions, 0.2  ton  per  lineal  foot. 

Total  lower  panel  fixed  load     ....     8.32;  tons. 
"      upper     "         "         "        .     .     .     .     3.375     " 

Total 11.700     " 

Lower  panel  load  at  end  of  arm       ...      5      tons. 
Upper      "         "     "    top    "   CP  ....     2.94    " 

The  trigonometric  quantities  required  will  not  be  re- 
peated. 

As  the  ends  of  the  span  are  simply  supported,  the  fixed 
load  stresses  are  those  existing  when  the  draw  is  open,  and, 
hence,  are  identical  with  those  found  for  the  fixed  load  in  Table 


Ij  =^1 


-I     -'    ^1         S'      5'      fil     s 


Y^  (mm)  r^ht^(^ 


Hi'    'i^  ^r-i-r    -i — -, — -' 

—14^9— 


-54:3:: 

Fig.  I. 


I  of  the  preceding  Article.  As  they  are  there  arranged  in 
convenient  shape  for  combination  with  the  moving  load 
.stresses  about  to  be  found,  they  will  not  be  reproduced  here. 


ENDS  SIMPLY  SUPPORTED.  1 6 1. 

The  moving  load  is  identical  with  that  used  in  Article  36, 
and  is  shown  by  the  diagram  on  preceding  page. 

For  the  same  reasons  given  in  that  Article,  the  system  of 
panel  concentrations  which  was  used  there  will  be  employed 
here,  instead  of  the  wheel  weights  shown  in  the  diagram. 
These  panel  concentrations  are  as  follows : 

Advance  load     ,        5.7    tons.    At  panel  i      .     25.55  tons. 

At  panel  2     .     .     1775     "  "       "      3     •     25.55     " 

"       "      4     •     .      19-35     " 

On  account  of  the  complete  continuity  at  the  center,  there 
will  be  no  condition  of  a  simple,  non-continuous  span  for  the 
moving  load  on  one  arm  only,  but  all  moving  load  stresses 
will  be  found  by  the  use  of  the  preceding  panel  concentra- 
tions in  connection  with  the  formulae  of  Article  35.  The 
reactions  R-^  and  R^  will  be  precisely  the  same  as  those  found 
on  page  146  of  Article  36.     They  are: 

Panel  load  Pat  panel  i,  R^  —  o.y \6\02  P ;  R^  =.  0.005443  P. 

"      2,  i?i  =  0.453178  P;  7^4  =  0.009526/1 

"      3,  R^  =  0.232204  P\  Ri  =  0.010886  P. 

"      4,  Ri  =  0.074153  P;  Ri  =  0.008165  P 

As  unbalanced  loads  will  be  considered  in  this  case,  these 
reactions  will  frequently  be  used  separately  in  the  succeeding 
computations;  but  in  those  special  circumstances  which  cause 
the  loads  to  be  balanced,  they  will  be  united  by  addition,  as 
was  done  in  the  preceding  Article. 

As  all  loads  produce  positive  or  upward  reactions  at  each 
end  of  the  bridge,  the  post  BPwiW  receive  its  greatest  stress 
when  the  moving  load  completely  covers  both  arms,  and  in 
such  a  manner  that  the  reaction  at  its  foot  will  be  the  great- 
est possible,  as  shown  by  the  values  of  the  panel  reactions 
given  above.     The  reaction  will  therefore  be : 

25.55  X  0.7161  =  18.297  tons. 
17.75  X  0.4532  ==    8.044     " 
25.55  X  0.2322  =     5.932     " 
19.35  X  0.0741  =     1.434     " 

Left  arm  total  =  33.707     " 
II 


-l62  WING  BRIDGES. 

19.35  X  0.0054  =  0.104  ton. 
25.55  X  0.0095  =  0.243     " 
17.75  X  0.0109  =  0.194     " 
25.55  X  0.0082  =  0.209    " 

Right  arm  total  =  0.750     " 

v.  {EP)  =  -  (33707  +  0.75)  sec  EP=  -  34.457  X  1.4 

=  —  48.24  tons. 

.*.  (i)  =  (2)  =  34-457  X  tanEP=  +  33.835  tons. 

As  Ti  cannot  sustain  any  compression,  the  hanger  T^q  will 
be  stressed  by  the  entire  load  at  its  foot,  unless  that  load 
added  to  the  coexistent  vertical  shear  in  the  upper  chord 
panel  a  exceeds  the  reaction  at  the  foot  of  EP.  Since  the 
latter  has  been  shown  to  be  34.457  tons,  while  the  maximum 
load  at  the  foot  of  T^q  is  only  25.55  tons,  that  excess  cannot 
exist,  but  ci  will  be  required  to  carry  over  shear  to  make  up 
the  reaction  found.     Hence: 

(7,o)  -=  +  25.55  tons. 

Since  the  function  of  c^  is  to  carry  over  shear  to  make  up 
the  reaction  at  the  foot  of  EP,  its  stress  will  be  the  greatest 
when  that  reaction  exceeds  by  the  greatest  amount  possible 
the  moving  load  between  it  and  c^,  added  to  the  coexistent 
shear  in  upper  panel  a.  As  the  advance  load  is  small  in  com- 
parison to  that  which  follows  it,  this  condition  will  exist  when 
the  moving  load  covers  both  arms,  with  the  advance  load  at 
the  foot  of  Zio-     The  reaction  at  the  foot  of  EP  will  then  be : 

5.7     X  0.7161  =    4.082  tons. 
25.55  X  0.4532  =  11.579 
17.75  X  0.2322  —    4.122 
25.55  X  0.0741  =     1.893 

Total  from  left  arm  ....     21.676 
Right  arm  reaction  as  above.        0.75 

Total  reaction      ....     22.426 

,  ,       22.426  X  20  —  5.7  X  21 
.'.  (^i)  —  — -"^ ■  sec  f  1  =  +  20  925  tons. 


ENDS   SIMPLY  SUPPORTED.  1 63 

The  advance  load  would  have  to  be  placed  at  the  foot  of 
Pg  if  it  were  sufficiently  large. 

At  the  end  of  the  Article  the  case  will  be  considered  in 
which  the  counter  c-^  is  omitted  and  the  members  7*4  and  P^ 
counterbraced.  It  will  then  be  shown  that  7^  and  Pj  can 
never  sustain  any  other  kinds  of  stresses  than  those  induced 
by  the  fixed  load.  Hence,  no  other  counter  than  c,  will  be 
required  in  the  present  instance,  and  the  main  web  stresses 
will  next  be  found. 

Since  the  function  of  the  main  web  members  is  to  carry 
load  or  shear  over  to  the  center,  one  condition  to  be  fulfilled  in 
order  that  they  may  receive  their  greatest  stresses  is,  in  gen- 
eral, to  make  the  reaction  at  the  end  of  the  arm  in  which  they 
are  located  as  small  as  possible.  As  the  main  web  stresses  in 
the  left  arm  are  now  sought,  therefore,  no  moving  load  must 
be  placed  on  the  right  arm  ;  because  all  such  load  increases 
the  upward  reaction  at  the  foot  of  EP. 

This  condition  holds  in  general  because  the  chords  of  draw 
spans  are  usually  either  parallel  or  so  inclined  to  each  other 
that  their  intersections  lie  witJioiit  the  span.  If,  however,  their 
intersections  lie  within  the  span,  the  reactions  at  the  free  end 
of  the  span  will  have  moments  of  the  same  sign  as  those  loads 
between  the  moment  origin  and  the  web  members  in  question. 
Hence  in  those  cases,  the  reactions  (with  moving  loads  omitted 
between  moment  origin  and  the  free  end)  should  be  as  large 
as  possible,  and  the  moving  load  should  cover  the  whole  of 
the  other  arm. 

The  intersection  of  the  panels  d  and  5  is  0.8  panel  length 
from  the  foot  oi EP,  and  0.2  panel  length  from  the  foot  of  Tiq; 
hence  the  entire  bridge  is  to  be  covered  with  moving  load 
for  the  greatest  stress  in  P^,  precisely  as  was  done  in  the  pre- 
ceding Article  for  the  same  member.     Therefore,  from  that 

Article : 

(/\)  =  -  56.875  tons. 

The  member  Ji  is  again  a  simple  hanger,  and  its  greatest 
stress  is  the  greatest  reaction  at  its  foot.     Hence; 

7i  =:  +  25.57  tons. 


1 64  SIVING  BRIDGES. 

For  the  remaining  web  members  the  moment  origin  is 
twenty  panel  lengths  from  the  end  of  the  span,  and  the  mo- 
ments of  the  reactions  will  have  signs  opposite  to  those  of 
the  loads.  No  moving  load  must,  therefore,  be  placed  on 
the  right  arm,  in  order  that  the  reactions  may  be  as  small 
as  possible. 

The  advance  load  will  be  placed  at  the  foot  of  T^  for  the 
greatest  stress  in  T^,  thus  making  the  reaction  at  the  foot 
oiEP: 

For  panel  i,  25.55  x  0.7161  =  18.295  tons. 

"         "     2,  17.75  ^  0.4532  =    8.04       " 

"     3.  25.55  ^  0.2322  =     5.933     " 

"          "     4,  5.7  X  0.0741  =    0.422     " 

Total  reaction  =  32.690     " 
^  25.55  X  21  +  17.75  X  22  +  25.55  X  23  -  32.69  X  20  ^^^  ^ 

23  ^ 

=  +  49.37  tons. 

The  greatest  stress  in  T^  requires  the  advance  load  at  the 
foot  of  T2,  giving  the  reaction  : 

For  panel  i,     17.75  ^  0.7161  =  12.71     tons. 
"      2,     25.55  X  0.4532  =  11.58       " 
"  "      3,        5.7     X  0.2322  =     1.325     " 

Total  reaction  =  25.615     " 

17.75  X  21  +  25.55  X  22  -  25.615  X  20 
•  •  W  3;  —  — — —  ^^  ^^^  ^  3 

=  +  2579» 
Since  P^  takes  its  greatest  stress  with  T^ : 

17.75  X  21  +  25.55  X  22  -  25.615  X  20 
^^^)-  23 

=  —  18.37  tons.. 


ENDS  SIMPLY   SUPPORTED.  1 65 

The  greatest  stress  in  7!,  requires  the  advance  load  to  be 
placed  at  the  foot  of  T-^.     Hence  the  reaction  will  be; 

For  panel  i,    25.55  ^  0.7 161  =  18.295  tons. 
"     2,      5.7     X  0.4532  =    2.585     " 

Total  reaction  =  20.880     " 

/-T^N  25.55      X     21     —     2O.8S      X     20  ™  ,       „    ^fL,     ^ 

.'.  (Ti)  =  -^-^^ sec  Ti=  +  7.765  tons. 

Since  P^  takes  it  greatest  stress  with  the  same  position  of 
moving  load  : 

/ox  25.55  X  21  —  20.88  X  20 

(Ps)  =  -  -^^ =  -  5.405  tons. 

In  order  to  illustrate  the  effect  of  moving  the  load  so  as 
to  bring  the  advance  concentration  of  5.7  tons  at  the  foot  of 
the  diagonal  whose  stress  is  sought,  let  the  stress  in  T^  be 
found  in  that  manner.  The  reaction  as  found  for  Z^  will  be 
20.88  tons.      Hence  : 

(7.)^  =  25.55x21+57^x22-20.88x20^^^  ^^ 

=  +  14.91  tons. 

This  result  is  only  a  little  more  than  one-half  the  greatest 
stress  in  T^  as  found  above.  If  the  advance  load  had  been 
large,  the  result  would  have  been  different. 

The  greatest  stresses  in  the  members  cc  and  LS  in  the  center 
panel  are  next  to  be  found.  These  members  carry  the  shear 
past  the  center  for  all  unbalanced  loads  on  either  arm  ;  hence 
the  stresses  in  them  will  occur  when  there  exists  the  greatest 
amount  of  unbalanced  load — i.e.,  when  the  moving  load  covers 
one  arm  while  the  other  arm  is  free  of  it.  This  also  follows 
clearly  from  the  fact,  demonstrated  in  Article  35,  that  all 
load  in  the  arm,  or  span,  /,  causes  ^3  to  be  negative  and  K2 
positive  ;  and,  hence,  since  the  stresses  in  a"  and  LS  result 
from    the    downward    pull    of   R^,  any  load    in    the  arm,  or 


1 66  SWING  BRIDGES. 

span,  4  would  reduce  the  negative  R^  by  balancing  some  load 
in  /i,  and  thus  correspondingly  reduce  the  stresses  in  question. 
Those  considei^tions  are  concomitant  with  complete  conti- 
nuity over  the  center. 

On  page  i6i  are  found  values  of  R\  and  R^  for  a  unit  panel 
load  at  each  panel  point  of  either  arm,  and  by  using  these  in 
Equation  {ya)  of  Article  35  the  following  values  of  R^  result, 
remembering  that  P  is  unity,  that  c  =  6.707,  and  that  n  has 
the  values  0.2,  0.4,  0.6,  and  0.8  respectively: 

For  panel  i,  R^—  —  0.6044 

"      2,  7^3=  -  1.0579 

"      3,  ^3  =  -  1-2095 

"      4,  Rs=  -  0.9076 

The  shear  in  the  center  panel  will  be  R^  +  R^  (a  numerical 
difference) ;  hence,  by  using  actual  panel  loads  : 

(—  0.6044  4-  0.0054)  X   19.35  =  —  11.59   tons. 
(—  1.0579  +  0.0095)  X  25.55  =  —  26.785     " 
(—  1.2095  +  0.0109)  X  17.75  =  ~  21.275     " 
(—  0.9076  +  0.0082)  X  25.55  =  —  22.98       " 

.-.    R^  +  R,=  -  82.630     '* 

Since  the  two  rods  cc  pull  against  opposite  ends  of  the  strut 
LS,  the  preceding  shear  of  82.63  tons  will  be  equally  divided 
between  them.     Hence: 

(cc)  =  41.315  X  sec  cc  =  +  57-715  tons. 

The  stress  in  LS  is  of  course  the  horizontal  component  of 
that  in  cc  ;  therefore  : 

(LS)  =  —  41.315  X  fan  cc  —  —  40.33  tons. 

The  preceding  calculations  show,  therefore,  that  the  total 
downward  reaction  to  be  supplied,  or  upward  pull  to  be  re- 
sisted, at  the  foot  of  each  cP  is  82.63  +  0.75  =83.38  tons; 
and  there  must  be  sufficient  weight  of  drum,  cross  beams, 
etc.,  as  well  as  strength  of  connections,  to  fulfill  the  require- 
ments of  that  condition. 


ENDS  SIMPLY  SUPPORTED.  1 6/ 

Since  the  center  panel,  or  span,  is  divided  into  two  stories, 
the  upper  half  of  cP  will  sustain  a  stress  different  from  that 
in  the  lower  half  in  consequence  of  the  action  of  the  two 
parallel  tension  braces  cc.  Again,  unbalanced  moving  loads 
will  produce  stresses  in  one  of  the  cP  posts  that  are  not  only 
different  in  amount  from  those  in  the  other,  but,  also,  different 
in  kind. 

It  has  just  been  shown  that  the  cc  tension  braces  inclined 
in  the  same  direction  as  /\  receive  their  greatest  stresses 
when  the  left  arm  is  entirely  covered  with  moving  load,  the 
right  arm  being  free  from  it.  It  necessarily  follows  that  the 
left  post  cP  receives  at  the  same  time  its  greatest  compres- 
sion. The  same  condition  of  loading,  it  has  also  been  shown, 
induces  the  greatest  possible  downward  reaction,  or  pull,  at 
the  foot  of  the  right  cP.  In  fact,  this  pull  produces  a  heavy 
tension  in  the  lower  story  of  that  cP  at  the  foot  of  which  the 
reaction  R^  exists  ;*but  in  the  upper  story  of  the  same  member 
a  small  tension  only  will  be  found,  due  to  the  small  moving 
load  compression  in  the  right-hand  member  d  induced  by  the 
reaction  R^.  It  will  be  further  apparent  that  the  moving  load 
on  one  arm  only  will  cause  the  greatest  compression  in  that 
cP  nearest  the  loaded  arm,  if  it  be  observed  that  the  member 
d  takes  its  greatest  tension  with  the  same  condition  of  load- 
ing;  because  if  the  other  arm  be  loaded,  the  upward  reaction 
at  the  end  of  the  arm  will  be  increased  by  the  amount  of  i?^, 
which  will  correspondingly  relieve  the  tension  in  d. 

Let  the  greatest  compression  in  the  lower  story  of  cP  be 
found  with  the  same  condition  of  moving  load  used  for  cc,  and 
in  doing  this  let  the  truss  be  supposed  to  be  divided  through 
d,  cP,  and  6  (the  cc  which  is  divided  does  not  act  and  is  there- 
fore neglected).  Moments  will  be  taken  about  the  intersec- 
tion of  d  and  6  in  the  left  arm,  by  the  reactions  R^  and  R^. 
Hence,  remembering  that  20.5  ~  27.5  =  0.7455  : 


Lower  half  {cP)  =  °-75  ><  9-9455  -  83.38  x  4-9455 

^  '  4.2 


=  —  96.4  tons. 


l68  SWING  BRIDGES. 

This  is,  of  course,  equal  to  the  vertical  component  of  the 
tensile  stress  in  d  added  to  the  vertical  components  of  the  two 
stresses  {cc).  It  will  presently  be  shown  that  the  tension  in  d 
is  37.875  tons;  hence: 

Lower  half  {cP^  —  37-875  x  0.364  +  2  x  41.315  =  —  96.4  tons. 

The  compression  in  the  upper  story  of  cP  will  be  simply 
that  in  the  lower  story  diminished  by  the  vertical  component 
of  (cc).     Hence : 

Upper  half  {cP)  —  —  (96.4  —  41.315)  =  —  55.085  tons. 

This  can,  again,  be  checked  by  moments  by  supposing  the 
truss  divided  through  d,  upper  half  cP,  LS,  cc,  and  6,  and  by 
including  the  moments  of  {LS)  and  {cc)  now  known  ;  but  it  is 
quite  unnecessary. 

Continuing  the  same  position  of  loading,  the  tension  in 
the  lower  half  of  the  other  ^/*  will  now  be  found.  Let  the 
truss  now  be  supposed  to  be  divided  through  the  right 
arm  d,  lower  story  of  right-hand  cP,  cc,  and  6;  then  let  mo- 
ments be  taken  about  the  intersection  of  6  and  d  in  the  right 
arm.  Remembering  that  ^4  =  0.75  ton,  and  R^  =  83.38 
tons,  while  the  lever  arm  of  {cc)  =  57-715  tons  is  4.2  panels 
divided  by  sec  for  cc  : 

T            u  If  ^       •      /  D\       0-75  X  0.8  +  83.38  X  4.2 
Lower  half  tension  (cP)  =  —^-^ ^^-^ -^— 

1^7.715  X  4.2 
_  3/  /   J ±_  __  ^  -2.21  tons. 

4.2  X  sec  cc 

This  result  can  be  checked  by  adding  the  vertical  compo- 
nent of  the  upper  {cc)  to  the  vertical  component  of  the  com- 
pression  now  existing  in  d  (in  consequence  of  the  upward 
reaction  ^^4),  whose  horizontal  component  will  presently  be 
shown  to  be  2.455.     Hence,  this  checking  operation  gives  : 

41.315  +  2  455  X  tan  f5  —  42.21  tons. 


ENDS   SIMPLY  SUPPORTED.  1 69 

The  upper  half  of  r/*  receives  only  the  vertical  component 
t)f  the  compression  in  d ;  hence  : 

Upper  half  tension  {cP')  =  2.455  ^  ^^'^  /^  =  +  0.895  tons. 

This  result  can  also  be  checked  by  moments  in  the  same 
general  manner  indicated  for  the  upper  half  of  the  other  cP, 
but  it  is  not  necessary. 

The  computations  for  the  web  stresses  are  thus  completed, 
leaving  those  for  the  chord  stresses  yet  to  be  shown. 

Unit  panel  moments,  as  described  on  page  152  of  Article  36, 
will  first  be  taken  about  the  foot  of  P.^  for  all  the  panel  points 
of  the  two  arms,  those  on  the  right  arm  being  represented 
by  Rjs!  i'ld  the  following  will  be  the  results : 

For  right  arm,  R^  x  2  =  +  zR^ 

"      panel  i,  0.7161  x  2  —  i  =  +  0.4322 

"  "      2,  0.4532  X  2  =4-  0.9064 

"  "      3,  0.2322  X  2  =  +  0.4644 

"  "      4,  0.0741   X  2  =  +  0.1482 

All  ihese  results  are  positive,  which  shows  that  a//  loads 
on  both  arms  produce  compression  in  d;  but  with  the  entire 
left  arm  loaded,  the  counter  Ci  comes  into  action  and  causes 
th'^  stress  in  a  to  be  the  same  as  that  in  d.  Therefore,  for  the 
greatest  compression  in  a  and  d,  moving  load  must  be  so 
placed  on  the  right  arm  as  to  give  ^4  its  greatest  value  of 
0.75  ton  (as  already  shown),  while  the  left  arm  is  entirely 
Tovered,  with  the  heavy  concentrations  at  panels  2  and  4. 
The  desired  moment  will  then  be: 

0.75  X  2  =1.5 

19.35  X  0.4322  =  8.363 
25.55  X  0.90641^  23.159 
17.75  X  0.4644=  8.243 
25.55  X  0.1482  =     3.786 

Total     .      .     .     45.051 

.  .        ,,.  45-051   X  27.5 

.,\  (a)  1=  (^)  =  —  tt£-^ <-^  sec  a  =  —  45.05 1 

V  /       V  /  29.333  ^^    5 

X  ^an  T^  X  sec  a  ^  —  42.31  tons. 


170  SWING  BRIDGES. 

Again,  for  the  chords  c  and  3,  unit  moments  about  the 
foot  of  P^  give  : 

For  R,,  ^4x3  +  3^4 

"    panel  i,  07161  x  3  -  2  =  +  0.1483 

"         "     2,  0.4532  +  3  -  I  =  +  0.3596 

*'         "     3,  0.2322  X  3  =  +  0.6966 

"         "     4,  0.0741   X  3  =  +  0.2223 

As  all  these  results  are  positive,  it  follows  that  all  panel 
loads  produce  compression  in  c  and  tension  in  3.  It  is  also 
evident  that  the  heavy  concentrations  must  be  placed  at 
panels  i  and  3.     Introducing  the  panel  concentrations: 

0.75  X  3  =  2.25 

25.55  X  0.1483  =  3.789 

1775  X  0.3596  =  6.383 

25.55  X  0.6966  =  17.798 

19.35  X  0.2223  =  4.301 


Total     .     .     .     34.521 
•••  (3)  =  ^"^'^^0666^^^  =  34.521  tail  T,=  +  30.965  tons. 

Also:  (<:)=  —  (3)  sec  a  =  —  31.005  tons. 

For  the  stresses  in  4  and  5  (equal  to  each  other),  the  unit 
moments  are  to  be  taken  about  the  top  of  /\ ;  they  are  as^ 
follows : 

For  Ri,  R4  X  4                 =  +  4R4 

"■    panel  i,  0.7161  x  4  —  3  =  —  0.1356 

"         "     2,  0.4532  X  4  —  2  =:  —  0.1872 

"         "     3,  0.2322  X  4  —  I  =  —  0.0712 

"         "     4,  0.0741  X  4          =  +  0.2964 

These  results  show  that  all  loads  on  the  right  arm  with 
that  at  panel  4,  only,  produce  tension  in  4  and  5.  Hence, 
the  moving  load  on  the  right  arm  is  to  remain  as  before, 
while  the  heavy  concentration  is  to  be  placed  at  the  foot  of 


ENDS  SIMPLY  SUPPORTED.  17 1 

Ti,  with  the  advance  load  of  5.7  tons  at  the  foot  of  P^.     By 
the  introduction  of  these  moving  loads : 

For  R^,  0.75  X  4  =  +    3 

"    panel  4,     25.55x0.2964    =+    7.573 


+  10.573 
3»  —  5'7     X  0.0712    =  —      ,406 


Total     .     .     .        +10.167 
(4)  =  (5)  =  10.167  X  tan  T.2=  -\-  8.735  tons. 

Since  the  loads  at  panel  points  i,  2,  and  3  produce  com- 
pression in  4  and  5,  the  greatest  compression  possible  must 
be  found.  It  will  result  from  placing  the  heavy  concentra- 
tion at  panel  point  2,  with  the  advance  load  of  5.7  tons  at 
panel  point  3.     The  actual  concentrations  then  give  : 

For  panel  point  i,  —  17.75  x  0.1356  =  —  2.407 
"  "  "  2,  —  25.55  X  0.1872  =— 4.783 
"         •'         "      3,     —    5.7     X  0.0712  =  —    .406 

Total     ...       =  —  7.596 
.•.  —  (4)  =  —  (5)  =  —  7.596  X  tan  T2=  —  6.525  tons. 

It  is  thus  seen  that  the  panels  4  and  5  are  the  only  por- 
tions of  the  lower  chord  that  can  ever  be  subjected  to  com- 
pression by  the  moving  load. 

In  order  to  determine  the  stress  in  d,  the  unit  moments 
must  be  taken  about  the  foot  of  cP,  and  they  are  : 

For  right  arm,  ^4x5                  =  +  5^4 

"  panel  point  i,  0.7161  x  5  —  4  =  —  0.4195 

'*         "         "  2,  0.4532  X  5  -  3  :=  -  0.7340 

"         "          "  3,  0.2322  X  5  —  2  =  —  0.8390 

"         "          "  4,  0.0741   X  5  —  I  =  —  0.6295 

Whence,  all  loads  on  the  right  arm  give  compression  to 
d,  and  they  must  be  removed  when  the  greatest  tension  in  d 


1/2  SWING  BRIDGES. 

is  sought.  The  greatest  moving  load  compression  in  d  will 
therefore  be : 

.{d)  =  —  0.75  X  5  X  -^-^  X  sec  ft  =  —  2.455  X  ^^^  ft 
42 

=  —  2.61  tons; 

and  since  under  this  loading  (on  the  right  arm  only)  the  rods 
cc  sloping  upward  to  the  left  do  not  act,  the  compression 
in  e  will  be  the  horizontal  component  in  d.     Hence: 

{e)  =  —  2.455  tons. 

If  all  the  load  on  the  right  arm  is  removed,  and  all  that  on 
the  left  arm  retained,  the  preceding  show  that  d  will  be  sub- 
jected to  its  greatest  tension.  By  placing  the  heavy  con- 
centrations at  panel  points  2  and  4,  and  introducing  all  the 
panel  concentrations  on  the  left  arm  in  the  unit  moments, 
there  will  result : 

For  panel  point  i,  19.35  x  0.4195  =    8.1 18 

"         "         "      2,  25.55  X  0.7340  =  18.753 

"         "         "      3,  17.75  X  0.8390=  14.892 

"         "         "      4,  25.55  X  0.6295  =  16.084 

Total     .     .     .     57.847 

.-.      {d)  =  57.847  X  ^^  -A  sec  ft  =  37.875  X  sec  ft 
42 

=  +40.3  tons. 

Under  this  condition  of  loading,  the  rods  cc  sloping  up- 
ward toward  the  right  do  not  come  into  action  ;  therefore 
the  compression  in  6  must  equal  the  horizontal  component 
of  the  tension  in  d\  or: 

(6)  =  —  37.875  tons. 

It  has  been  shown  that  moving  load  on  one  arm,  only,  pro- 
duces compression  in  e  ;  hence,  to  obtain  the  greatest  tension 


ENDS   SIMPLY  SUPPORTED. 


^71 


in  that  member,  the  whole  of  both  arms  must  be  loaded  with 
balanced  moving  loads,  which  is  precisely  the  condition  used 
in  Article  36,  on  page  155,  where  there  was  found: 

{e)  —  +  35.42  tons. 

This  completes  the  computations  for  all  the  moving  load 
stresses,  and  Table  I  shows  them  grouped  so  that  they  may 
be  combined  with  those  due  to  the  fixed  load  as  found  in 
Article  36. 

TABLE  I. 


Member. 

Stress. 

Member. 

Stress. 

T, 

+  25. 57      tons. 

c 

—  31.005  tons. 

T, 

+  49 

37 

d 

\  +  40 

3 

T, 

+  25 

79       " 

"(-    2 

61 

T, 

+     7 

765     " 

' 

\  +35 

42       " 

r.o 

+  25 

55       " 

"/-    2 

455     " 

Ci 

+  20 

925     " 

I 

+  33 

S35     " 

EP 

-48 

24       " 

2 

+  33 

835     •' 

P^ 

-56 

875     '• 

3 

+  30 

965     " 

P^ 

-  18 

37       " 

S  +     8 

735     " 

P. 

-    5 

405     " 

4 

'(-    6 

525     " 

Upper  cP 

(  +    0 
/  -  55 

895     " 
085     •' 

5 

\  +     8 
/-    6 

735     " 
525     •' 

Lower  cP 

i  +  42 

21 

6 

-  37 

875     '• 

/-96 

40       " 

cc 

+  57 

715    ;] 

a 

-  42 

31 

LS 

-  40 

33 

b 

—  42 

31 

By  comparison  of  these  results  with  those  of  Table  I  on  page 
155,  it  is  observed  that  the  assumption  of  complete  continuity 
produces  web  stresses  either  slightly  greater  in  every  instance 
(except  those  due  to  the  moving  load  on  one  arm  only)  than 
the  hypothesis  of  partial  continuity,  or  identical  with  it.  The 
chord  stresses  are  also  usually  slightly  greater,  or  equal,  ex- 
cepting, also,  those  due  to  moving  load  on  one  arm  only  in 
the  case  of  partial  continuity,  which  are  materially  greater. 
Toward  and  at  the  ends  of  the  span,  therefore,  the  assump- 
tion of  a  simple  non-continous  span  gives  much  larger  stresses, 
both  in  chords  and  web,  than  are  found  under  the  supposition  of 
continuity,  as  would  be  anticipated.  With  these  exceptions, 
however,  the  differences  are  unimportant,  and  either  method 


174  SWING  BRIDGES. 

could  be  used  with  indifference,  and  that  selected  which 
would  result  in  the  least  labor  of  computation. 

An  enormous  difference  is  seen,  however,  in  the  magnitude 
of  the  stresses  both  for  the  center  posts  and  central  diago- 
nals, and  in  the  labor  of  their  computation,  cc  and  LS  suffer 
no  stress  under  partial  continuity,  but  are  subjected  to  heavy 
stresses  if  the  continuity  is  complete.  The  posts  cP  also  suf- 
fer very  heavy  compression  in  the  latter  case,  and  a  compar- 
itively  light  compression  in  the  former,  due  only  to  the  fixed 
load.  The  lower  stories  of  the  same  members  are  also  sub- 
jected to  heavy  tension  when  the  continuity  is  complete.  In 
the  present  instance  this  tension,  added  to  the  vertical  com- 
ponent of  the  stress  in  the  lower  cc,  is  42.21  +  41.315  —  83-525 
tons;  and  since  the  fixed  load  compression  in  <:/^  is,  by  Table 
I  of  Article  36,  36.785  tons,  the  total  maximum  upward  pull 
to  be  resisted  at  the  foot  of  cP'xs: 

83.525  -  36.785  =  4^-74  tons. 

Now  the  lower  chord  fixed  load  at  the  foot  of  cP  is 
24  X  0.3025  =  7.26  tons,  and  the  weight  of  the  drum,  etc., 
which  may  be  assumed  to  be  concentrated  at  the  same  point, 
is  not  more  than  9  or  10  tons ;  or,  say,  9.74  tons.  Hence  the 
amount  of  anchorage  which  would  have  to  be  provided  under 
the  assumption  of  complete  continuity  is  : 

46.74  —  (7.26  +  9.74)  =  29.74  tons  ; 

and  unless  this  were  supplied,  each  foot  of  r/' would,  in  turn, 
rise  and  fall  with  varying  conditions  of  moving  load,  so  that 
not  only  very  destructive  hammering  would  take  place,  but, 
also,  the  condition  of  those  spans  continuous  over  two  sup- 
ports would  be  displaced  by  that  of  two  unequal  spans  con- 
tinuous over  one  support,  thus  vitiating  the  computations 
underh'ing  the  design  of  the  trusses. 

The  realization  of  the  conditions  requisite  for  the  case  of 
complete  continuity,  therefore,  involves  considerably  in- 
creased difficulty  and  expense  in  connection  with  the  design 
of  the  center  portion. 


ENDS  SIMPLY  SUPPORTED. 


175 


Inasmuch  as  the  main  truss  members  have  been  found  to 
sustain  essentially  the  same  stresses  in  either  case,  except 
where  those  under  partial  continuity  largely  exceed  the  others, 
and  inasmuch  as  all  draw-bridge  formulae  drawn  from  the 
theory  of  the  continuous  beam  involve  the  uniformity  of 
the  moment  of  inertia  of  all  normal  sections,  thereby  incur- 
ring a  very  considerable  error  in  at  least  some  of  the  resulting 
computations,  it  is  much  more  rational  and  in  harmony  with 
better  judgment  to  use  the  methods  of  partial  continuity  in 
all  cases.  These  and  similar  considerations  have  led  engi- 
neers to  almost  universally  adopt  the  partial-continuity  as- 
sumption in  the  construction  of  swing  bridges. 

It  is  to  be  observed  that  the  unbalanced  uplift  at  the  foot 
of  cP  and  the  stresses  in  cc  and  LS  will  increase  rapidly  as  the 
length  of  the  center  panel  or  span  decreases ;  and,  hence,  that 
it  is  advisable  to  make  that  center  panel  as  long  as  possible. 
The  method  of  computation  will  be  in  no  way  changed  if 
the  center  panel  is  designed  with  one  storj^  only,  instead  of 
two. 

By  combining  the  fixed  load  stresses  in  Table  I  of  Article 
36  with  those  due  to  the  moving  load  given  in  Table  I  of  this 
Article,  Table  II,  giving  the  resultant  stresses  in  all  the  mem- 
bers, at  once  results. 

TABLE    II 


Member. 

Tension. 

Compression. 

Member. 

Tension. 

Compression. 

Ti 

Tons 

+  33-895 
+  99-39 

-\-  62  605 

+  30.335 
+  17-41 
+  20.925 
4-     7.0 

Tons 

b 
c 
d 
e 
I 
2 
3 
4 
5 
6 
cc 
LS 

Tons 
•f     20.37 
+     44-98 
+  13Q.23 
-f-  128.40 
+    28.925 
+     13-485 

Tons 
-    21.94 

7'o 

Ti 

Ti 

Tio 

—       8.14 

-  4.91 

-  20.35 

-  44-93 

-  84.045 

-  84  045 

-  130.855 

EP 

-  41.24 

-  80.605 

-  47-965 

-  24.495 

-  91-87 

-  133-85 

-  37-39 

Pi 

Pi 

Upper  cP 
Lower  cP 

+     5-425 
+     4-92 

+     57-715 

-    40-33 

a 

1/6  SWING  BRIDGES. 

These  resultant  stresses  are  not  very  different  from  those  in 
Table  II  of  page  156,  except  in  those  members  already  indi- 
cated. In  this  case,  upper  chord  c  and  lower  chord  3  do  not 
need  counterbracing,  while  they  required  such  treatment  in 
the  case  of  partial  continuity.  In  actual  practice,  however, 
the  entire  lower  chord,  and  the  upper  chord  panels  a,  b,  and  c, 
would  all  be  designed  with  compression  cross  sections — i.e., 
they  would  be  counterbraced.  The  lower  story  of  ^P  must 
also  be  counterbraced,  and  ordinary  construction  would  make 
it  so  even  if  it  were  not  required. 

The  existence  of  the  central  diagonals  cc  designed  to  trans- 
fer shear,  results  in  some  ambiguity  in  the  upper  chord  stresses 
at  the  center.  As  the  moving  load  passes  on  the  bridge,  one 
pair  of  those  diagonals  receive  their  greatest  stresses,  and 
then  they  may  either  be  supposed  to  be  relieved  with  the 
further  progress  of  the  train,  or  they  may  be  supposed  to  hold 
essentially  their  greatest  stresses  while  the  other  pair  gradu- 
ally take  the  same  condition,  as  the  moving  load  covers  the 
entire  structure.  Either  supposition  fulfills  the  requisites  for 
equilibrium,  but  the  former  will  give  the  greatest  possible 
stress  in  e  and  was  used  in  the  preceding  computations,  be- 
cause the  stress  in  the  same  upper  chord  member  will  be 
decreased  under  the'  latter  supposition  by  an  amount  corre- 
sponding to  the  horizontal  components  of  the  stresses  in  the 
diagonals  cc.  This  condition  of  ambiguity  cannot  be  avoided 
under  the  assumption  of  complete  continuity,  but  disappears 
if  the  continuity  is  assumed  to  be  partial  only.  The  proper 
method,  therefore,  is  to  compute  the  greatest  possible  stress 
in  each  member,  and  use  it  in  the  design,  and  this  has  been 
done  in  the  present  case. 

The  reaction  R^  and  R2  at  the  ends  and  adjacent  to  the 
center,  respectively,  remain  to  be  written,  but  they  can  be 
taken  directly  from  results  already  found  in  the  preceding 
Article.  For  this  purpose,  and  for  the  reasons  fully  explained 
in  Article  36,  the  moving  load  will  be  taken  as  uniform  and 
at  0.775  to'""  P^i"  lineal  foot  for  each  truss.  The  greatest  end 
reaction  will  exist  with  the  moving  load  over  the  entire  struct- 
ure, and  it  has  been  shown  on  page  1 57  that  its  amount  will  be  : 


ENDS  SIMPLY  SUPPORTED.  177 

21   ■^12 

^1  =  32.173  +  —  +6=  48.829  tons. 

The  half  panel  load  is  that  adjacent  to  the  end,  whicK 
does  not  affect^  the  trusses  but  forms  a  part  of  the  reaction 
to  be  supported  by  the  truss  ends,  while  the  6  tons  is  the 
wheel  concentration.  Since  the  moving  load  on  each  arm 
produces  a  negative  or  downward  reaction  at  the  opposite 
side  of  the  drum,  the  greatest  reaction  R2  will  be  produced 
with  the  moving  load  on  the  adjacent  arm  only.  The  reac- 
tion R^  due  to  the  four  panel  loads  at  the  panel  points  1,2,  3, 
and  4  will  be  : 

i?,'  =  (0.7 161  4-  0.4532  +  0.2322  +  0.0741)  =  31.448  tons. 

Hence: 
7^2'  =  4  X  21.312  —  31.448  ■] —  X  21.312  +  6  =  78.399  tons. 

The  third  term  in  the  second  member  is  for  the  half  panel 
adjacent  to  the  center  and  the  central  panel,  while  the  6 
tons  is  for  the  driving-wheel  concentration,  as  already  ex- 
plained. As  shown  on  page  158,  the  total  fixed  load  at  the 
foot  of  Pi  is  62  tons;  hence  the  total  reaction  desired  is: 

7?2  =  78.399  +  62  =  140.399  tons. 

Tkc  omission  of  counters. 

If  all  counters  are  omitted,  it  will  usually  be  necessary 
to  counterbrace  some  of  the  main  web  members.  In  the 
present  case,  T^  will  be  most  in  need  of  such  treatment.  The 
position  of  moving  load  required  to  give  its  maximum  com- 
pression to  7^4  is  the  same  as  that  used  in  finding  the  great- 
est tension  in  c^  on  page  162,  and  the  corresponding  reactioa 
at  the  foot  of  EP  was  there  found  to  be  22.426  tons. 

Hence : 

,^.            22.426  X  20—  5.7  X  21         ™ 
—  (7^)  = ^-^ —sec  7^  =  —  21.47  tons. 


178  SWING  BRIDGES. 

But  it  has  been  shown  that  the  fixed  load  tension  in  T^  is 
22.57  tons.  No  counterbracing,  therefore,  according  to  these 
results,  is  required.  But  the  tension  excess  is  so  small  that 
the  member  should  be  made  with  a  compression  section, 
otherwise  the  moving  load  shock  might  overcome  the  excess 
and  cause  the  member  to  buckle.  A  rule  of  good  practice  is 
to  counterbrace  if  at  least  one-fourth  of  the  computed  mov- 
ing load  stress  added  to  itself  overcomes  that  due  to  the 
fixed  load,  and  perhaps  a  better  rule  is  to  add  one-third. 

With  the  same  position  of  loading  there  will  result: 

,„.       22.426  X  20  —  5.7  X  21 

(P3)  =  -^ ^^     ^  ' ==  +  14.95  tons. 

If  one-fourth  of  14.95  tons  be  added  to  itself,  the  result 
will  be  less  than  19.09  tons,  the  fixed  load  compression  in  P^\ 
hence  no  reversion  of  stress  can  take  place,  although  the 
character  of  the  pin  connection  would  enable  very  consider- 
able tension  to  be  resisted. 

For  the  greatest  compression  in  Z3,  and  tension  in  /*2»  the 
advance  load  of  5.7  tons  must  be  placed  at  the  foot  of  /g. 
The  reaction  at  the  foot  of  EP  will  then  be  : 

5.7  X  0.4532  =  2.583 
25.55  X  0.2322  =  5.932 
17.75  X  0.0741  ~  1.316 


Total  from  left  arm     .     .     .     9.831  tons. 
Right  arm  reaction      .     .     .     0.75       " 

Total     .     .     .     10.581     '\ 

Hence : 

,„.,            10.581   X  20  —  5.7  X  22  ^  ^  ^^^  ^ 

—  (T;)  = 2 ^-J. sec  2^3  =  —  5.265  tons. 


22 


And 


,„v        10.581   X  20  —  5.7  X  22 

(^2)  =  — ^ ^,     ^^ =  +  3.75  tons. 


TABLES  AND  DIAGRAMS. 


179 


Both  of  these  results  are  so  small  in  comparison  with  the 
■opposite  fixed  load  stresses  that  no  counterbracing  is  re- 
quired. 

These  computations  show  that  even  if  the  counter  c^  is 
omitted,  no  counterbracing  will  be  required,  except  in  the 
case  of  7"^,  as  explained. 

Article  37a. — Tables  and  Diagrams. — Turntables   and  Engines. 

The  labor  of  stress  computations  by  the  methods  given  in 
the  three  preceding  Articles  can  be  reduced  to  a  very  small 
amount  by  the  aid  of  the  tables  and  diagrams  which  follow. 
They  are  devised  for  the  purpose  of  showing  the  reactions  at 

TABLE    I. 


either  extremity  of  a  three-span,  rim-bearing  drawbridge  for 
a  square  crossing  and  for  arms  of  equal  length.  The  ratios, 
€,    of   either    arm    to    the    center    span   or    panel,   as    shown 


i8o 


SWING  BRIDGES. 


immediately  preceding  Equation  (3)  of  Article  35,  are  given- 
in  Table  I,  with  the  greatest,  least,  and  mean  values,  taken 
from  fifteen  drawbridges  as  they  have  actually  been  designed. 
They  cover,  therefore,  a  range  that  will  include  nearly  all 
practical  cases  of  pin-connected  structures.  The  columns  of 
the  table  show  the  reactions  R^  and  R^  for  a  unit  panel  load 
placed  successively  at  distances  from  the  free  end  of  the  arm 
/[  which  vary  by  .05/].  These  reactions  due  to  unit  panel 
loads  are  computed  from  Equations  (5^)  and  (6«)  of  Article 


\ 

\ 

V 

\ 

\ 

> 

V 

'^ 

\ 

M 

^? 

0 

* 

y-. 

03 

^d 

^ 

' 

« 

*^ 

'*' 

^ 

N 

\ 

\ 

k 

1 

1 

^, 

S; 

k 

V 

^ 

<^ 

Ss 

1 

J 

"^ 

55=1 

.4  .5 

Fig.   I. 


35,  with  the  values  of  c  given  at  the  heads  of  the  columns  of 
the  table.  As  an  example,  if  the  unit  load  be  placed  at  the 
distance  0.2/1  from  the  free  end  of  the  arm,  the  reaction  7?, 
will  have  the  values  0.71355  and  0.72112  for  the  values  of 
I  ^  U  =  c  =  8.822  and  4.395  respectively.  The  corresponding 
reactions,  ^4,  will  be  .004401  and  .007311.  The  actual  reac- 
tions for  any  panel  load  will  be  found  by  simply  multiplying 
those  given  in  the  table  by  the  actual  panel  load  in  pounds 
or  tons,  as  the  case  may  be. 

It  will    be    observed  that    the    reactions    vary    very   little 


TABLES  AND  DIAGRAMS. 


I8l 


between  the  limits  of  range  of  the  values  of  c.  Hence,  for 
values  materially  outside  of  those  limits,  the  corresponding 
reactions  may  be  assigned  by  the  aid  of  Figures  i  and  2  with 
sufficient  accuracy.  Those  figures  exhibit  the  results  given 
in  Table  I.    Figure  i  gives  the  reactions  7?,,  and  Figure  2  the 


ibO 

/ 

'' 

N. 

140 

/ 

/ 

s 

y 

130 

/ 

\ 

/ 

\ 

120 

1 

/ 

110 

¥ 

f 

/ 

y 

s, 

s. 

y 

100 

V 

/ 

s 

V 

\ 

/ 

/ 

\ 

\ 

90 

/ 

1°/ 

\ 

\ 

"0 

/ 

/ 

^ 

— 

\ 

\ 

80 

/ 

/ 

/ 

^' 

s 

Ik 

^ 

\ 

a 

0 

/ 

/ 

#, 

/ 

\ 

\ 

70 

« 

1 

/ 

%V" 

\ 

\ 

/ 

/  i 

V 

60 

// 

1 

// 

50 

\  1 

/ 

I 

/ 

\ 

40 

I  u 

\\ 

, 

7 

/ 

30 

/ 

// 

A 

/ 

20 

// 

f 

i 

II 

ul 

10 

1 

\l 

\ 

J 

i  .5 

Fig.  2. 


reactions  R^.  The  three  curves  so  nearly  coincide  in  Figure  i 
that  only  those  for  the  two  values  of  r  =  8.822  and  c  =  4.395 
are  shown.  For  practical  working,  that  figure  should  be 
drawn  to  at  least  double  the  size  shown.  In  Figure  i  the 
tabular  values  of  the  reactions  R^  have  been  taken  full 
size,  while  those  for  R^,  in   Figure  2,  have  been  multiplied 


1 82  SWING  BRIDGES. 

by  io,ooa  The  figures  will  enable  either  of  the  reactions  to 
be  read  at  a  glance  for  any  length  of  panel  in  any  length  of 
arm.  They,  as  well  as  Table  I.,  are,  therefore,  perfectly  gen- 
eral for  a  wide  range  of  the  value  of  c. 

After  the  reactions  are  read  from  the  table  or  the  diagrams, 
moments  for  either  the  web  or  chord  stresses  can  readily  be 
written  ;  and  from  these  moments  the  stresses  themselves 
will  at  once  result  in  accordance  with  the  methods  of  the 
preceding  Articles.  If  the  chords  are  parallel,  moments  will 
not  be  required  for  the  web  stresses,  as  the  latter  can  be  at 
once  written  from  the  shears. 

In  many  cases,  particularly  if  the  chords  are  parallel,  the 
values  given  in  Tables  II.  to  VIII.  will  be  found  very  conven- 
ient for  seven-  to  nineteen-panel  drawbridges,  in  which  the 
center  panel  is  equal  in  length  to  the  others.  They  have 
been  computed  by  Mr.  Frank  C.  Osborn,  consulting  engineer, 
of  Cleveland,  Ohio,  for  his  own  practice,  who  has  kindly  given 
the  author  the  privilege  of  using  them  in  this  connection. 
They  express  the  shears  and  moments  on  the  basis  of  each 
panel  load  being  unity,  and  of  each  panel  being  one  unit  in 
length.  Each  actual  shear  will  therefore  be  found  by  multi- 
plying each  tabular  shear  by  the  actual  panel  load,  and  each 
actual  moment  will  be  found  by  multiplying  each  tabular 
moment  by  the  product  of  the  actual  panel  load  by  the  actual 
panel  length.  These  tables  also  show  the  greatest  shears  and 
moments  at  the  various  panel  points.  The  reactions  due  to 
the  various  panel  loads,  by  the  aid  of  which  the  shears  and 
moments  are  obtained,  are  computed  from  Equations  (5^) 
and  (6«)  of  Article  35,  except  those  which  belong  to  the 
simple  spans,  which,  of  course,  follow  the  law  of  the  lever. 
If  the  chords  are  not  parallel,  the  web  stresses  must  be 
determined  by  the  method  of  moments,  as  illustrated  in  the 
two  preceding  Articles.  Even  if  the  length  of  the  center 
panel  should  differ  to  some  extent  from  that  of  the  others, 
Table  I.  and  Figs,  i  and  2  show  that  the  values  in  Tables  II. 
to  VIII.  will  not  be  sensibly  changed. 


TABLES  AND   DIAGRAMS. 


183 


TABLE    II. 
(BCD  E  .F  G 


i         J         ^ 

A         ALL  EQUAL        A 


f^  7   PANELS  1^ 

Note. — Shear  in  panel  ab  =■  reaction  at  a,  and  Shear  cd  =  reaction  at  d. 


Loads  at : 

Shear  in  Panel : 

Moment  at 

ab 

he 

cd 

b 

c 

d 

b  and  g 

+  0  568 
+  0.210 

-0.432 
+  0.210 

-0.432 
-0.790 

+  0.568 
+  0.210 

+  0.136 
+  0.420 

—0.296 
—0.370 

c  and  f. 

Maximum ■ 

As  a  Simple  Span - 

+  0.778 
+ I .000 

+  0.210 
-0.432 

+  0-333 
-0-333 

—  1.222 

—  1. 000 

+  0.778 
+ I. 000 

+  0.556 
+ I. 000 

-o!666 

TABLE    III. 

(B C  D  E  F  G 


b  2  rf 

9  PANELS 


tL r 


f 

A 


S 


A  9  PANELS  l'^  A  ALL  EQUAL 

Note. — Shear  in  panel  ah  =  reaction  at  a,  and  Shear  de  =  reaction  at  g. 


Loads  at; 

Shear  in 

Panel 

Moment  at : 

ab 

be 

cd 

de 

b 

c 

d 

e 

i  and  i 

+  0.665 
+  0.364 
+  0. 131 

-0-335 
+  0.364 
+  0.131 

-0.335 
—0.636 
+  0.131 

-0-335 
—0.636 
—0.869 

+  0.665 
+  0.364 
+  0.131 

+0.330 
+  0.728 
+0.262 

—0.005 
+  0.092 
+0.393 

-0.340 
-0.544 
-0  476 

Maximum 

As  a  Simple  Span - 

+  1.160 
+  1.500 

+  0.495 
-0-335 

+  0.750 
-0.250 

+  0.131 
-0.971 

+  0.250 
-0.750 

-1.840 
—  1.500 

+  1.160 

+  1.500 

+ I . 320 
+  2.000 

+  0.485 
—0.005 

+  1.500 

—  1.360 

184 


SWING  BRIDGES. 


TABLE  IV. 


A  A 


Ji  t  k  T 

ALL  EQUAL 


A  I  I     PANELS 

Note. — Shear  in  panel  ab  =  reaction  at  a,  and  Shear  cf  =  reaction  aty. 


Loads  at  : 

Shear  in  Panel : 

Moment  at  : 

ab 

be 

cd 

de 

e/ 

b 

c 

d 

e 

/ 

+  0.726 
+  0.471 
+  0.252 
+  0.089 

-0.274 
+  0.471 
+  0.252 
+  0.089 

-0.274 
-0.529 

+  0.252 
+  0.089 

-0.274 
-0.529 
-0.748 
+  0.089 

-0.274 
-0.529 
-0.748 
— 0.911 

+  0.726 
+  0.471 
+  0.252 
+  0.089 

+  0.452 
+  0.942 
+  0.505 
+  0.178 

+  0.178 
+  0.412 
+  0.757 
•+0.268 

+  1.615 

-0.095 
— C.117 
+  o.oog 
+  0-357 

-0.369 
—0.646 
-0.738 
-0-554 

c  and  k 

Maximum  . . . .  - 

+  1.538 
+  2.000 

+  0.812 
-0.274 

+  1.200 
—0.200 

+  0.341 
-0.803 

+  0.600 
—0.600 

+  0.089 
-1. 551 

+  0.200 
-1.200 

—  2.462 
—2.000 

+  1.538 

+  2.077 

+  0.366 
—0.212 

+  2.000 

-2.307 

As    a    Simple  j 
Span 1 

+  2.000 

+  3.000 

+  3.000 

TABLE  V. 
B      C      D       E      F     G H       I      K      L      M    N 


b       c       3"  e      / 
13  PANELS 


m         n 
ALL  EQUAL 


A  13  PANELS      A    A      ALL  EOUAL       A 

Note. — Shear  in  panel  ab  —  reaction  at  a,  and  Shear  ^  =  reaction  at^. 


Loads  at  : 

Shear  in  Panel : 

ab 

be 

cd 

de 

'f 

fS 

+  0.768 
+  0.548 
+  0.350 
+  0.185 
+  0.065 

—0.232 

+  0.548 
+  0.350 
+  0.185 
+  0 . 065 

—0.232 

-0.452 

+  0.350 
+0.185 
+  0.065 

—0.232 

—0.452 
—0.650 
+  0.185 
+  0.065 

—0.232 
-0.452 
—0.650 
—0.815 
+  0.065 

-o.23« 

-0.452 
-0.650 
-0.815 

-0-93S 

j 
—  I 

Maximum 

As  a  Simple  Span 

+ 1.916 
+  2.500 

+  1.148 
—  0.232 

+ I . 667 
-0.167 

+  0.600 

-0.684 

+  1.000 

—0.500 

+  0.250 
-'•334 

+  0.500 
—  1. 000 

+  0.065 
-2.149 

+  0.167 
-1.667 

-3  084 
—  2.500 

TABLES  AND  DIAGRAMS. 


loi 


Loads  at  : 

Moment  at  : 

b 

c 

d 

e 

/ 

S 

b  and  n 

+0.768 
+0.548 
+0.350 
+0.185 
+0.065 

+  0-537 
+  1.097 
+  0.700 
+  0.370 
+  0.129 

+  0.305 
+  0.645 
+ 1.050 
+  0.556 
+  0.194 

+  0.074 
+  0.193 
+  0.400 
+  0.741 
+  0.259 

—0.158 
-0.259 
-0.250 
-0.074 
+  0.323 

-0.390 
—0.710 

Maximum \ 

+ I. 916 
+2.500 

+  2.833 
+  4.000 

+  2.750 
+  4.500 

+  1.667 
+  4.000 

+  0.323 
—0.741 

+  2.500 

-3.501 

TABLE    VI. 
B     CDEFGHIKLMNOP 


A 


y   J   /! 


I       III       )i        op        q 
A 


A  15   PANELS  A      A  ALL  EQUAL 

Note. — Shear  in  panel  ab  =  reaction  at  <?.  and  Shear  gh  =  reaction  at  h. 


Loads  at  : 


b  and  / 

c  and  o 

d  and  n 

e  and  m 

_/"and  / +0. 142 

^and  ^ +0.049 


Ma.ximum. 


As    a    S  i  m  pie 
Span 


+0.799 
+0.606 
+  0.427 
+0.270 


+  2.293 


Shear  in  Panel : 


—0.201 
+  0.606 
+0.427 
+  0.270 
+  0.142 
+  0.049 


+  1.494 
—0.201 


+  2.143 
-0-143 


cd 


—0.201 
-0.394 
+  0.427 
+  0.270 
+  0. 142 
+  0.049 


-0.888 
-0.595 


+  1.429 
-0.429 


de 


—0.201 
-0.394 
-0-573 
+  0.270 
+  0.142 
+  0.049 


+0.461 
-1.168 

+  0.857 
-0.857 


ef 


—0.201 
-0-394 
-0-573 
-0.730 
+  0.142 
+  o . 049 


+  0.191 
—  1.898 


+  0.429 
-1.429 


fg 


—0.201 
-0.394 
-0.573 
-0.730 
-0.858 
+  o . 049 


+  0.049 

-2.756 


+0.143 
-2.143 


gh 


201 
394 
573 
730 


Loads  at  ; 

Moment  at  : 

b 

c 

d 

e 

/ 

g 

h 

b  and  ii 

+  0.799 
+  0.606 
+0.427 
+  0.270 
+0.142 
+0.049 

+  0.600 
+  1.212 
+  0.855 
+0.540 
+  0.283 
+0.098 

+  0.398 
+  0.819 
+  1.282 
+  0.811 
+  0.425 
+0.148 

+  0.198 
+  0.425 
+  0.709 
+  1.081 
+  0.566 
+  0.197 

—0.003 
+  0.031 
+  0.137 
+  0.351 
+  0.708 
+  0.246 

—0.203 
-0-363 
-0.436 
-0.379 
—0.150 
+  0.295 

-0.404 

-0-7S7 

—  1.009 

—  1. 109 

—  1.009 
—0.656 

d  and  n 

e  and  th 

/and  /...:. 

.f  and  k 

Maximum - 

As  a  Simple  Span.. 

+  2.293 
+  3.000 

+  3-588 
+  5.000 

+  3.883 
+  6.000 

+  3.176 
+  6.000 

+  1.470 
—0.003 

+  5.000 

+  0.295 
-1-531 

+  3.000 

-4-944 

1 86 


SWING  BRIDGES. 


A 


TABLE   VII. 
B     C     D     E     F     G     H     I     K     L     M     N     6     P     Q     ff 


K 


bcdejghik       I      m      n      o      p       q      r       i 
A  17   PANELS  A      A  ALL  EQUAL  A 


Note. — Shear  in  panel  ab  —  reaction  at  a,  and  Shear  /^z  =  reaction  at  i. 


Loads  at  : 


b  and  r 

c  and  g 

</and  / 

e  and  o  

/  and  « 

g  and  »/ 

h  and  /. 

Maximum \ 

As  a  Simple  Span ■! 


Shear  in  Panel : 


cd 


+  0.8231  —0.177  —0.177 

+  0.651  +0.651;  —0.349 

+  0.489  +0.489  +0.489 

+  0.342  +0.342I  +0.342 

+  0.215  +0.215I  +0.215 

+  0.112  +0.112  +0.112 

+  0.039  +0.C39'  +0.039 


+  2.671 


+  3-500 


+  1. 


+  1.197 
—0.526 


+  2.625     +I.875 
-0.125    -0.375 


-0.177 
-0.349 
— 0.511 
+  0.342 
+  0.215 
+  0.112 
+  0.039 


C/  fg 


-0.177 
-0.349 
— 0.511 
-0.658 
+  0.215 
+  0.112 
+  o . 039 


hi 


+  0.708  +0.366 

-1.037  -1-695 

+  1.250  +0.750 

—  o . 750  —  1 . 250 


-0.177I 
-0.349 

— 0.511 

—0.658 

-0.7851 

+  O.H2 
+  0.039 


-0.177 
-0.349 
— 0.5II 
-0.658 
-0.785 
-0.888 

+  o . 039 


-0.177 
-0.349 

— 0.5II 
—0.658 

-0.78s 

-0.88& 
—0.961 


+  0.151     +0.039 

—2.480    -3-368    —4.329 


+  0.375 
-1.875 


+  0.125 
—2.625 


Loads  at: 


h  and  r 

c  and  g 

d  and  / 

e  and  o 

yand  n 

jf  and  m 

h  and / 

Maximum - 

As  a  Simple  Span 


Moment  at  : 


+0.823 
+  0.651 
+  0.489 
+  0.342 
+  0.215 
+  0.112 
+  o . 039 


+  2.671 


+  0.646I  +0.470 

+  1.303!  +0.954 

+  0.979  +1.468 

+  0.684  +1.026 

+  0.430  +0.644 

+  0.224  +0.336 

+  0.077  +0.116 


+  4-343 


+  3 . 500    +  6 . 000 


+  5.014 
+  7.500 


+  0.293 
+  0.605 
+  0.9S7: 
+  1.369 
+  0.859 
+  0.448 
+  0.155 


+  4.686 


+  8.000 

I  I 


+  0.116 
+0.257 
+  0.447 
+  0.711 

+  1.074 

+  0.559 
+  0.193 


+  3-357 
+  7.500 


—0.061 
—0.092 
—0.064 
+  0.053 
+  0.289 
+  0.671 
+  0.232 


+  1.245 
—0.217 


—0.441 
-0.575 
—0.605 

-0.497 
— 0.217 
+  0.271 


+  0.271 
-2-573 


+  3.500 


-0.41S 
-0.789 
-1.086 
— 1.263 

—  1.282 

—  1.105 
—0.691 


-6.631 


TABLES  AND  DIAGRAMS. 


l%7 


h 


TABLE  VIII. 

:B  •  C^D^e     F     G      HI      K     L     M    N     O      P     Q     R     S      T 


a.      b       C      a       »      f      g       k      i       k       I      m       n      o       p      q        r       s       t      iT 
A  19   PANELS  f^       ^  ALL  EQUAL  A 


Note. — Shear  in  panel  ai  =  reaction  at  <?,  and  Shear  ik  =  reaction  at  k. 


Loads  at : 

Shear  in  Panel : 

ai 

6c 

cd 

de 

ef 

fs 

gh 

hi 

ik 

^and  i 

c  and  J 

</and  r 

e  and  g 

/andp 

g-  and  o 

A  and  n 

«■  and  m 

+  0.842 
+  0.687 
+  0.540 
+  0.403 
+  0.280 
+  0.175 
+  0.091 
+  0.031 

—0.158 
+  0.687 
+  0.540 
+  0.403 
+  0.280 
+  0.175 
+  0.091 
+  0.031 

—0.158 

-0.313 

+0.540 

+  0.403 
+0.280 
+0.175 

+  0.091 
+  0.031 

-0.158 
-0.313 
—0.460 
+  0.403 
+  0.280 
+  0.175 
+  0.091 
+  0.031 

-0.158 

-0.313 
—0.460 

-0.597 
+  0.280 
+  0.175 
+  0.091 
+  0.031 

-0.158 
-0.313 
—0.460 
-0.597 
—0.720 
+  0.175 
+  0.091 
+  0.031 

-0.158 

-o.3'3 
—0.460 
-0.597 
-0.720 
-0.825 
+  0.091 
+  0.031 

+  0.122 
— 3°73 

+  0.333 
-2.333 

-0.158 
-0.313 
—0.460 

-0.597 
—0.720 
-0.825 
-0.909 
+  0.031 

-0.158 

-0.313 
—0.460 
-0.597 
—0.720 
—0.825 
-0.909 
—0.969 

Maximum...  < 

As  a  Simple  1 
Span.        1 

+  3.049 
+  4.000 

+  2.207,   +1.520 
—0.158    —0.471 

+  3. Ill    +2.333 
— o.iii    —0.333 

+  0.980 
-0.931 

+ 1.667 
—0.667 

+  0.577 
-1.528 

+  1.111 
— i.iii 

+  0.297 
—  2.248 

+  0.667 
-1.667 

+  0.031 
-3.982 

+  0.111 
-3. Ill 

-4-951 
—4.000 

Loads  at 

Moment  at  : 

b 

c 

d             e 

/ 

sr 

h 

i 

k 

*and  t 

c  and  s  

d  and  r 

e  and  g 

/and/ 

g  and  0 

h  and  n 

/and  m 

+  0.842 
+  0.687 
+  0.540 
+  0.403 
+  0.280 
+  0.175 
+  0.091 
+  0.031 

+0.684 

+  1.374 
+  1.080 
+  0.806 
+0.560 
+0.350 
+0.182 
+  0.062 

+  0.526    +0.3681   +0.210 
+  1.061     +0.748    +0.435 
+  1.620    +1.160J   +0.700 
+  1.209    +1.612    +1.015 
+  0.840    +1.120!   +1.400 
+0.525    +0.700    +0.875 
+  0.273    +0.364'   +0.455 
+  0.093    +0.1241    +0.155 

+0.052 
+0.122 
+0.240 
+0.418 
+0.680 
+ 1 . 050 
+0  546 
+0.186 

—0.106 
— 0.191 
—0.220 
-0.179 
—0.040 
+  0.225 
+  0.637 
+  0.217 

—0.264 
-0.504 
—0.680 
-0.776 
—0.760 
—0.600 
-0.272 
+  0.248 

-0.423 
-0.814 
-1. 142 

-•.37s 

—  1.482 

—  I  428 

—  1.184 
-0.719 

Ma-ximum...  \ 

As  a  Simple  1 
Span t" 

+  3.049 
+  4.000 

+  5.098 
+  7.000 

+  6.147    +6.196 
+  9.000'  +  10.000 

+  5.245 
+ 10.000 

+  3-294 
+  9.000 

+  1.079 
-0.736 

+  7.000 

+  0.248 
-3.856 

+  4.000 

-8^567 

The  drums  of  rim-bearing  turntables  should  be  of  sufficient 
depth  to  prevent  upward  deflection  from  materially  disturb- 
ing a  uniform  distribution    of  load  over  the  rollers.      The 


I^ya  SWING  BRIDGES. 

upward  pressure  of  the  latter  constitutes  an  upward  loading 
on  the  lower  flange  of  the  drum,  and  the  points  on  the 
upper  flange  of  the  latter,  at  which  the  truss  load  is  applied, 
form  the  supporting  points  of  the  continuous  drum  girder. 
The  loading  on  the  lower  drum  flange  should  be  as  nearly- 
uniform  as  possible,  and  in  order  to  secure  that  result,  the 
drum  depth  should  be  as  great  as  possible.  It  is  also  neces- 
sary that  the  truss  load  should  be  carried  to  the  upper  drum 
flange  at  as  many  equidistant  points  as  may  be  found  prac- 
ticable. In  long  and  heavy  draw  spans  it  will  be  necessary 
to  carry  the  truss  loads  to  the  drum  through  a  combination 
of  transverse  and  longitudinal  girders,  in  order  to  secure  the 
requisite  number  of  points  of  application  on  the  upper  flange. 
A  calculation  of  the  girder  strength  of  the  drum  can  be  made 
by  assuming  that  its  segments  are  beams  with  a  span  length 
equal  to  the  distance  between  points  of  application  of  the 
truss  loads  on  the  upper  flange,  and  that  they  are  loaded 
with  the  uniform  roller  pressure.  Although  the  drum  is  con- 
tinuous, these  beams  should  be  considered  non-continuous, 
for  they  are  not  straight,  and  a  failure  to  secure  the  assumed 
uniformity  of  loading  may  essentially  destroy  the  advantages 
of  continuity.  The  results  of  all  such  computations  must, 
however,  be  strongly  tempered  with  those  of  experience. 
The  drum  section  must  be  such  as  to  avoid  any  appreciable 
deflection  ;  its  depth  should  never  be  less  than  one-third  the 
distance  between  adjacent  points  of  support  on  the  upper 
flange,  and  one-half  is  better  practice.  If  the  total  truss 
load  does  not  require  nearly  all  the  rollers  which  the  circum- 
ference of  the  drum  affords,  the  rule  may  be  proportionately 
modified,  but  not  otherwise. 

The  Thames  River  bridge,  carrying  about  2,400,000  pounds 
dead  load  on  the  rollers,  has  eight  equidistant  points  of  sup- 
port on  its  drum,  with  32  feet  diameter  and  5  feet  depth. 

The  500  feet  single-track  Arthur  Kill  draw  span  has  about 
1,450,000  pounds  of  dead  load  resting  at  eight  equidistant 
points  on  a  drum  27^  feet  in  diameter  and  3^  feet  deep. 

The  Central  bridge  across  the  Harlem  River  at  New  York 
has  two  concentric   drums  44  feet  and   36  feet   in  diameter 


DRUMS  AND  ENGINES.  1^7^ 

with  sixteen  points  of  support  on  each.  The  total  truss 
weight  is  about  3,800,000  pounds,  and  the  depth  of  the  drum 
is  5  feet.  The  total  truss  weight  of  the  Kingsbridge  Road 
bridge  across  the  Harlem  Ship  Canal  at  New  York  is  about 
1,500,000  pounds,  and  it  is  carried  at  twelve  equidistant  points 
of  support  on  a  drum  39  feet  in  diameter  and  5  feet  deep. 

It  is  usually  very  easy  to  secure  eight  equidistant  points  of 
support  on  the  drum  of  an  ordinary  single-track  draw  span, 
with  a  depth  of  drum  of  at  least  30  to  36  inches,  and  such 
an  arrangement  should  be  required.  The  resulting  distri- 
bution of  load  on  the  rollers  will  be  found  very  satisfactory. 

The  power  required  to  be  exerted  by  an  engine  to  turn  a 
drawbridge  is  expended  in  three  parts.  One  portion  is  used, 
at  the  beginning  of  the  operation  of  opening  or  closing,  in 
developing  the  maximum  velocity  possible,  and  is  stored  for 
a  short  time  as  the  actual  energy  of  the  structure  in  motion  ; 
it  subsequently  performs  work  against  some  brake  arrange- 
ment by  which  the  bridge  is  brought  to  rest.  A  second  por- 
tion is  used  in  performing  work  against  the  entire  frictional 
resistance  of  the  moving  parts  of  the  structure  and  machin- 
ery ;  while  the  third  and  last  portion  is  required  to  overcome 
the  wind  resistance  when  the  wind  blows  against  one  arm 
with  a  total  pressure  which  is  not  balanced  by  that  against  the 
other.  The  rolling  friction,  or,  rather,  the  entire  friction,  of  a 
drawbridge  in  motion,  has  been  determined  by  Mr.  Theodore 
Cooper  for  the  Second  Avenue  double-track  railroad  bridge 
over  the  Harlem  River  at  New  York;  and  for  the  Thames 
River  double-track  railroad  bridge  at  New  London,  Conn.,  by 
Messrs.  Boiler  and  Schumacher;  and  the  results  of  these  in- 
vestigations can  be  found  in  the  Transactions  of  the  Ameri- 
can Society  of  Civil  Engineers,  for  December,  1891.  The 
total  moving  weight  of  the  Second  Avenue  bridge  was  880,- 
000  pounds,  and  that  of  the  Thames  River  bridge  2,400,000 
pounds.  The  length  of  the  latter  is  500  feet,  and  the  diame- 
ter of  the  drum  32  feet.  Mr.  Cooper  found  the  coefificient 
of  frictional  resistance  for  the  Second  Avenue  bridge  to  be 
.0038 — i.e.,  a  force  of  3.8  pounds  per  i,ooo  pounds  of  total 
weight  moved  would  have  to  be  applied  tangentially  at  the 


iS^C  SWING  BRIDGES. 

center  line  of  the  track  (or  to  the  drum)  in  order  to  overcome 
the  total  friction. 

Messrs.  Boiler  and  Schumacher  found  a  coefficient  of 
about  .004  for  the  Thames  River  bridge,  or  4  pounds  per 
1,000  pounds  moved.  The  greater  part  of  the  "total  fric- 
tion "  is  the  rolling  friction  at  the  drum.  These  two  instances 
are  the  most  valuable  rim-bearing  drawbridge  investigations 
of  the  kind  ever  made  in  this  country,  and  as  the  workman- 
ship, fitting  of  track,  etc.,  were  of  an  unusually  excellent 
character,  it  is  probably  advisable  to  take  the  coefficient  of 
friction  for  ordinary  draw  spans  at  .01,  or  10  pounds  per  1,000 
pounds  of  weight  moved. 

In  the  Transactions  of  the  American  Society  of  Civil 
Engineers  for  1874,  Mr.  C.  Shaler  Smith  gave  the  results  of 
a  number  of  less  complete  but  very  interesting  tests  of  rim- 
bearing  draw  spans  in  about  the  ordinary  conditions  of  work- 
manship and  running  order.  He  found  the  total  friction  to 
vary  from  4  to  8  pounds  per  i,ooo  pounds  of  weight  moved. 

The  power  required  to  give  the  desired  velocity  of  rotation 
to  a  drawbridge  will  depend  upon  the  time  allowed  for  open- 
ing or  closing.  Draws  operated  by  power  are  usually- 
opened,  or  closed,  in  one  to  three  minutes.  Small  draws 
operated  by  hand  will  consume  three  to  eight  minutes. 

If  drawbridges  are  operated  against  an  unbalanced  wind 
pressure,  the  necessary  power  increases  very  rapidly.  Seven- 
eighths,  or  even  nine-tenths,  of  the  total  capacity  of  a  well- 
proportioned  drawbridge  engine  may  be  exerted  against  wind 
pressure  when  the  structure  is  moved  in  a  moderately  high 
wind.  A  comparatively  small  amount  of  power  is  required 
to  overcome  the  friction. 

In  the  500  feet  double-track  Thames  River  bridge,  with 
moving  parts  weighing  about  2,400,000  pounds,  not  more 
than  5  or  6  horsepower,  at  most,  was  expended  in  develop- 
ing the  acceleration  and  overcoming  the  total  friction.  An 
unbalanced  wind  pressure  of  5  pounds  per  square  foot  on  one 
arm  would  have  required  only  a  little  less  than  30  horsepower 
to  turn  the  draw  against  it,  or  double  that  amount  for  a 
lO-pound  wind. 


DRUMS  AND   ENGINES.  iSjd 

The  computation  of  the  work  required  to  turn  a  draw- 
bridge will  require  its  moment  of  inertia  to  be  taken  about  a 
vertical  axis  through  the  center  of  the  drum.  It  will  be  suf- 
ficient for  this  purpose  to  consider  the  trusses,  lateral  brac- 
ing, floor  system,  and  track  as  a  homogeneous  prism  with 
length  /  and  width  zv.  This  portion  of  the  weight  is,  for  all 
practical  purposes,  five-sixths  the  total  moving  weight  W^for 
single-track  railroad  spans,  and  seven-eighths  the  same  total 
weight  for  double-track  spans.  Hence  the  moment  of  inertia 
/'  of  this  portion  of  the  weight  will  be — if  ^  is  the  approxi- 
mate constant,  32.2,  for  gravity — for  single  track  spans  : 

^,  ^  I  wjw'  +  n  ^  5  iv{w'  +  n        . 

Or  for  double-track  spans: 

9^g  ^' 

The  moment  of  inertia  of  the  drum,  rollers,  etc.,  can  be 
considered  concentrated  at  the  distance  R  =  radius  of  the 
drum,  from  the  axis.  Hence  the  moment  of  inertia,  /",  of 
this  portion  of  the  weight  will  be,  for  single-track  spans : 

'"  =  ^ <'>• 

Or,  for  double-track  spans  : 

r-^     (4 

Hence  the  total  moment  of  inertia  will  be  : 

f=/'  +  r (5). 

The  nominal  horsepowers  of  engines  fitted  to  a  number  of 
railroad  and   heavy  city  draw  spans  which   have  proved  to 


iSje 


SWING  BRIDGES. 


be  very  satisfactorily  operated   are  given  in   the  tabulation 
below  : 


Bridge. 


500  feet  double  track  railway 

400  feet  city  bridge 

270  feet  city  bridge , 

400  feet  double  track  railway 
300  feet  double  track  railway 
362  feet  single  track  railway 
217  feet  double  track  railway 


Weight  of 
moving  parts. 

Engines. 

LBS. 

H.P. 

2,400.000 
4,200,  COO 
I, 800,000 

40 

50 
40 

2,000,000 

35 

1.250.000 
684,000 
600,000 

to  to  to 

00c 

The  tendency  has  been  toward  an  increase  of  engine 
power,  in  consequence  of  some  of  the  earlier  and  smaller 
engines  having  shown  insufificient  capacity' in  winds  and 
other  contingencies  of  drawbridge  operation. 

The  preceding  considerations  apply  to  rim-bearing  turn- 
tables only,  which  are  now  exclusively  used  for  all  draw 
spans  over  a  length  sufficiently  great  to  require  the  structure 
to  be  of  the  "  through  "  type. 

The  center-  or  pin-bearing  type,  in  which  the  entire 
moving  weight  of  the  structure  is  carried  on  a  center  pin 
or  pivot,  is  used  for  short  spans  only.  This  center  pin 
or  pivot  may  vary  from  eight  to  twelve  or  more  inches 
in  diameter,  so  that  the  pressure  per  square  inch  of  bear- 
ing surface  will  not  take  greater  values  than  two  thousand 
pounds  to  twenty-five  hundred  pounds.  The  center  pin, 
of  wrought  iron,  or,  preferably,  steel,  frequently  rests  upon 
a  flat  disk  of  phosphor-bronze  in  order  to  reduce  friction 
and  wear.  The  bearing  faces  of  the  pin  and  disks  should 
be  channeled  or  grooved  to  allow  entrance  for  a  heavy 
lubricant. 

In  the  Transactions  of  the  American  Society  of  Civil  En- 
gineers for  1874,  Mr.  C.  Shaler  Smith  gave,  as  the  result 
of  a  number  of  tests  of  center-bearing  draw  spans,  the  fric- 
tional  resistance,  if  exerted  at  the  circumference  of  the  center 
pin,  at  i-fo  of  the  total  weight  turned.     If   W^  is  that  weight 


DRUMS  AND   ENGINES.  1 87/ 

in  pounds,  and  F  the   force  of  friction  supposed  exerted  in 
the  circumference  of  the  pin,  then  : 

F=.og  W (6). 

If,  again,  /  is  time  in  minutes  required  by  one  man  to  open 
the  draw,  or  close  it,  and  if  ^  is  the  diameter  of  the  center 
pin  in  feet,  he  gave  : 

/  z=  .09  H^^ "^ (7). 

4  X   10,000  ^  ^ 

The  time  given  by  Equation  (7)  will  usually  be  insufficient 
for  the  requirements  of  one  man,  even  if  no  other  considera- 
tion than  that  of  friction  be  involved. 

In  the  contingency  of  an  unbalanced  wind  pressure  on  one 
arm  of  the  draw,  one  man  may  not  be  able  to  put  the  bridge 
in  motion,  nor  to  hold  it  against  the  wind.  Hence  none  but 
the  smallest  draws  in  unimportant  situations  are  made  de- 
pendent on  one  man  power. 

On  account  of  the  intermittent  character  of  the  operations 
of  the  motive  power  of  a  draw-bridge,  gas  or  hot  air  en- 
gines are  admirably  adapted  to  that  purpose.  Their  increased 
economy  and  essential  cessation  of  expenses  when  the  bridge 
is  not  being  turned  meet  the  requirements  of  the  situation  in 
a  very  satisfactory  manner.  As  far  as  they  have  been  tried 
they  leave  little  to  be  desired. 


Art.  38. — Ends  Simply  Resting  on  Supports— One  Support  at  Centre — 

Example. 

The  general  principles  fundamentally  involved  in  this  case 
are  not  different  from  those  of  the  two  preceding  ones,  except 
in  the  number  of  points  of  support  at  the  first  pier.  All  the 
fixed  load  of  the  bridge  is  carried  to  the  central  point  of  sup- 


188  SWING  BRIDGES. 

port,  whether  the  bridge  is  open  or  closed  ;  the  end  supports 
furnish  reactions  for  the  moving  load  only. 

The  truss  to  be  taken  as  an  example  is  the  one  shown  in 
the  accompanying  figure,  in  which  the  arms  are  of  equal 
length. 

The  general  formulae  to  be  used  for  the  reactions  at  A,  B, 
and  C,  and  for  the  bending  moment  at  the  centre,  are  equa- 
tions (ii),  (12),  (13),  and  (10)  respectively  of  Art.  35.  These 
equations  may  be  written  as  follows,  remembering  that  /^  = 
/a  =  /,  and  M.^  =  M: 


^=-ip\-^''^'' 


^)z+ip{i'-z')z\ 


(I). 


R,  =  j^hp{l-a)+M\   ........     (2). 

R,=  j\^hPa+hp,~2M'^ (3). 

R,=  'j^^hp{l-z)+M'^ (4). 


It   is  to  be  remembered  that  c  is  measured  from  A  or  (7, 
according  as  the  left  or  right  arm  is  considered. 


The  following  are  the  data  to  be  used  : 

Total  length  =  AC=2l=2AB=  2BC  =  144  feet. 
Uniform  depth  =  dD  —  bB  —  16  feet. 
Panel  length  —  AD  =  DE  =  etc.  =  13  feet. 
BH  =  BH'  =  7  feet. 


EXDS  SIMPLY  RESTING   ON  SUPPORTS. 


189 


Total  fixed  weight  per  foot  =  1,200  pounds  (nearly). 

Upper  chord  panel  fixed  weight  =  W  —    2.73  tons. 
Lower      "         "         "  "       =  W  =   5.00     " 

Uniform  panel  moving  load        =  w   =  19.50     " 

The  moving  load  traverses  the  lower  chord,  and  the  weight 
of  the  floor  system  is  taken  at  nearly  350  pounds  per  foot. 

On  account  of  the  extra  weight  of  the  locking  apparatus, 
the  fixed  weight  at  ^,  or  C,  will  be  taken  at  3  tons,  and  will 
be  denoted  by  u\. 

As  is  clear  from  the  figure,  all  inclined  web  members,  ex- 
cept the  end  posts,  are  for  tension  only,  while  the  verticals 
are  compression  members. 

As  the  ends  A  and  C  are  neither  latched  down  nor  lifted 
up,  either  arm  is  a  single  truss  simply  supported  at  each  end, 
for  all  moving  loads  which  rest  upon  it,  so  long  as  there  are  no 
movmg  loads  on  the  other  arjn. 

For  exactly  the  same  reasons,  therefore,  as  those  given  in 
the  preceding  Article,  any  counter,  as  dE,  will  sustain  its 
greatest  stress  when  the  moving  load  extends  from  its  foot  to 
the  centre,  if  no  other  moving  load  rests  on  the  bridge. 

It  must  still  be  borne  in  mind  that  in  connection  with  any 
counter  stress,  the  stress  in  the  vertical  which  cuts  its  upper 
extremity  is  to  be  found,  for  such  a  one  may  be  the  greatest 
stress  in  the  vertical. 

Again,  resume  the  general  expression  for  the  shear  in  any 
web  member : 

s—  S  —  n(W  ^-  W)  —  n'w ; 

in  which  S  is  the  shear  at  one  extremity  of  the  arm,  and  n 
and  n'  the  numbers  of  fixed  and  moving  panel  weights  re- 
spectively between  the  same  end  of  the  arm  and  the  web 
member  in  question.  In  considering  the  main  web  members, 
5  will  be  taken  adjacent  to  the  centre,  and,  in  the  present 
example,  at  an  indefinitely  short  distance  from  B  in  the  arm 
AB. 

For  a  given  condition  of  loading  in  AB,  it  is  evident  that 
the  smaller  is  ^1  the  greater  will  be  6".     But  Eq.  (i)  shows 


I  no  SWING  BRIDGES. 

1 
that  M  \s  always  negative.     Hence  so  long  as  '2P{1  —  z)  re- 
mains the   same,   Eq.  (2)  shows  that  R^  decreases  as  M  in- 
creases (numerically). 

Again,  Eq.  (i)  shows  that  J/ will  have  its  greatest  numerical 

2 
value,  other  things   remaining  the  same,  when  ^P{P  —  z^)z 

has  the  greatest  value  possible ;  i.  c,  when  the  moving  load 

covers  the  whole  of  the  arm  BC.     With  a  given  value,  there- 

1 
fore,  for  ^P{1  —  z),  R^  will  be  the  least  possible  when   the 

moving  load  covers  the  whole  of  the  other  arm,  or  the  whole 
of  BC;  consequently  5  will  be  the  greatest  under  the  same 
conditions.  Now  having  found  under  what  circumstances  S 
is  the  greatest,  precisely  the  same  reasoning  used  in  the  pre- 
ceding Articles  shows  that  s  will  be  the  greatest,  under  the 
same  circumstances,  when  ;/  is  zero. 

Any  inclined  main  wch  vieinber,  then,  will  sustain  its  greatest 
tensile  stress  when  the  moving  load  extends  from  its  foot  to  the 
free  extremity  of  the  arm  in  which  it  is  found,  and  covers  at  the 
same  time  the  zvJiole  of  the  other  arm. 

A  few  main  web  stresses  in  all  trusses  of  this  case  are  a 
little  singular  in  character,  but  are  no  exceptions  to  this  rule. 
Those  for  the  example  taken  will  be  noticed  in  the  proper 
place. 

Any  vertical  web  member,  unless  acting  as  a  counter,  will 
sustain  its  greatest  compression  in  connection  with  the  great- 
est tension  in  the  inclined  main  web  member  which  cuts  its 
upper  extremity. 

In  seeking  the  greatest  main  web  stresses,  it  may  happen 
that  the  reaction  Ry  becomes  zero ;  this,  however,  changes 
nothing  in  the  method. 

Since  either  arm  is  a  simple  truss  for  all  moving  loads  rest- 
ing on  it  (supposing  none  on  the  other),  every  such  load  tends 
to  cause  the  same  kind  of  stress  throughout  the  same  chord. 
Consequently,  as  in  the  previous  Article,  the  greatest  tension  in 
the  lower  chord,  and  compression  in  the  upper,  will  exist  when  the 
moving  load  covers  one  arm  only.  These  stresses  will  be  found 
in  that  portion  of  the  arm  adjacent  to  its  free  extremity. 

The   greatest   chord   stresses   of   the    same   kind  as  those 


E.VDS   SIMPLY  RESTING   ON  SUPPORTS. 


191 


caused  by  the  fixed  load,  can  be  found  with  the  least  labor 
by  first  determining  all  the  stresses  due  to  t/ie  fixed  load  alone, 
and  tabulating  them. 

The  stresses  caused  by  the  moving  load  alone  are  then  to 
be  determined  by  the  aid  of  the  following  considerations. 

Let  that  moment  be  considered  negative  which  causes 
tension  in  the  upper  chord  and  compression  in  the  lower. 
All  moments,  then,  caused  by  the  fixed  or  moving  loads  are 
negative,  and  all  those  produced  by  the  upward  reactions  R-^ 
are  positive.  Now  the  compression  in  any  lower  chord  panel 
may  be  found  by  taking  moments  (such  will  be  negative  if 
compression  exists)  about  the  panel  point  vertically  over  that 
extremity  nearest  the  centre.  The  general  expression  for 
such  compression  will  be : 

R{np  —  n'wt 


in  which  n  is  the  number  of  the  panel  from  the  free  extremity 
of  the  arm,  n  the  number  of  the  moving  panel  loads  on  the 
arm,  /  the  distance  of  the  centre  of  gravity  of  the  moving 
loads  71  w  from  the  origin  of  moments,  /  the  panel  length,  and 
d  the  depth  of  the  truss. 

The  numerator  of  this  expression  must  be  negative,  and  it 
is  desired  to  find  what  value  of  n  will  give  it  its  greatest 
negative  value. 

Now  since  every  panel  moving  load  on  the  arm  AB 
increases  R^  (as  a  positive  quantity),  it  appears  from  the 
figure  that  71  must  not  be  greater  than  («  —  i),  and,  farther,  it 
must  belong  to  loads  between  the  panel  considered  and  the 
free  extremity  of  the  arm.  Since,  however,  /  varies  with  71 
the  above  expression  may  have  its  greatest  negative  value 
when  71  is  less  than  (;/  —  i). 

These  considerations  are  independent  of  the  general  charac- 
ter of  R^\  it  has  already  been  seen,  however,  that,  with  a 
given  loading  on  AB,  R^  will  be  the  least  when  the  moving 
load  covers  the  whole  of  BC  also.  Hence  in  order  to  find  the 
greatest  tension   in  the  upper  chord  and  compression  in  the 


192 


SWING  BRIDGES. 


lower  due  to  the  moving  load,  the  method  of  procedure  is  as 
follows : 

Throughout  the  whole  operation  the  moving  load  is  to 
entirely  cover  one  arm,  as  BC.  The  moving  load  is  then  to 
cover  the  other  arm  from  the  free  extremity  to  any  panel 
point,  and  the  stresses  in  the  panels  situated  between 
the  end  of  the  train  and  the  centre  are  to  be  computed. 
This  operation  is  to  be  repeated  for  every  panel  in  that  arm 
not  wholly  covered  by  the  moving  load.  From  these  results 
the  greatest  stresses  may  be  selected  and  then  added  to  the 
fixed  load  stresses. 

The  character  of  these  operations  and  the  reasons  for  them 
will  be  much  more  evident  after  the  example  is  treated. 

The  following  values  will  be  needed,  and  depend  only  on 
the  data  already  given. 


^=  13 
z  —  26 

^=39 
^=52 

^  =  65 


/-^=  59 
/  -  ^  =  46 

^--  =  33 
I  —  z  —  20 
I  —  z  =    7 


(/2  —  ^)^  =  65,195.00 
(/■^  —  s^)z  =  1 17,208.00 
{P  -  z')z  =  142,857.00 
(/'^  —  z^)z  =  128,960.00 
{P  -  z^)z  =    62,335.00 


Angle  AdD  =  a. 
"       HbB  =  fi. 


tan  a 
tan  fi 


0.8125. 
043  75- 


sec  a  =  1.29. 
sec  yS  =  1 .09. 


4/2       20736 
P=  19.5  tons  =  w ; 


=  0.00004823 

P 

I 


0.271 


The  following  values  of  M  are  found  by  substituting  the 
proper  numerical  values  in  Eq.  (i).  The  moving  load  is  taken 
to  cover  the  whole  of  BC  and  so  much  oi  AC  d^^  is  indicated 
by  the  values  of  z. 


ENDS  SIMPLY  NESTING   ON  SUPPORTS. 


193 


In  arm  AB  \  2  =      13 


M=  —  547.00 


((     <(       <( 


((     ((       « 


«     <<       (t 


;  ^  = 

r3 
[26  • 

.  M=  —  657.00 

(13 

;  ^r  =  ■<>  26  . 

39 

.  .  M=  —  792.00 

;  ^=  - 

13 
26 

39  •  ' 

152 

.  M=  —913.00 

;  ^=  " 

13 
26 

39  •  • 

52 

.  M=  —  972.00 

165 

The  following  values  may  now  be  written.  Those  of  7?i 
are  found  by  simply  substituting  the  proper  numerical  quanti- 
ties in  Eq.  (2) ;  the  moving  load,  as  the  values  of  M  show,  is 
taken  to  cover  the  whole  of  BC. 


2=      13 


p  i 


-  :2  {I—  z)  =  15.99  tons  .  .  R^—    8.39  tons. 
=  28.46     "      .  .  R,=  19.33     - 


13 

26 

39 


3740 


R^  -  26.40    " 


13 
26 

39 

152 


42.82     " 


^1  =  30.14    " 


13 
26 

39 

52 

.65 

13 


=  44.72     "     .  .  i?i  =  31.22     " 


IQ4  SWING  BRIDGES. 

The  stresses  in  the  counters  will  first  be  sought,  i.e.,  those 
in  the  arm  AB. 

As  a  trial  let  the  moving  load  cover  the  points  F,  G,  and 
H,  and,  as  before,  let  R^  be  the  general  expression  for  the  re- 
action at  A.     Hence: 

y?i  =  3  X  19.5  X  -^ '-  =  16.25  tons. 

72 

Since  3  -}-  2  x  7.73  >  R^,  2P  =  o  at  the  panel  point  E,  and 
no  counter  is  needed  between  e  and  F. 

Moving  load  over  EH. 

26  1; 
7?i  =  4  X  19.5  X    ^^  =  28.71  tons. 
72 

As  3  +  2  X  7.73  +  19.5  >  Rxy  '2P=  o  at  E,  and  dE  is  the 
first  and  only  counter  needed. 

The  vertical  component  of  the  stress  in  dE  is 

s  =  R^-T,  +  7.73  ==  17.98  tons. 

Hence,        {dE)  =  17.98  x  sec  a  =  +  23.19  tons. 

The  greatest  compression  in  the  end  post  Ad,  and  tension 
in  the  vertical  d'D  will  exist  when  the  moving  load  covers  the 
whole  of  the  arm  AB,  the  other  carrying  none,  and  with  such 

loading : 

•3  •? 

y?l  zz:    5    X   19.5    X    -^  =  44-69  tons. 

72 

Hence,         (Ad)  =  -  {R^  -  t,)  x  see  a  =  -  53.78  tons. 
At  the  same  time : 

(d'D)  =  +  {19.5  +  5.00)  =  +  24.5  tons. 

The  stresses  in  the  main  web  members  are  next  to  be  de- 
termined. 

Those  in  Ad  and  d'D  will  occur  under  circumstances  to  be 
indicated  hereafter. 


E.VDS  SIMPLY  RESTIXG   OX  SUPPORTS. 


19! 


The  following  operations  are  in  accordance  with  the  princi- 
ples already  shown. 

Moving  load  on  BC  and  at  D. 
R^  =  8.39  tons ;  hence,  for  the  shear  in  De: 

J=  -  /?i  +  (3  +  7.73  +  19-5)  =  21.84  tons. 

Hence,  (^^)  =  s  x  seca  =  +  2S.iy  tons. 

Also,  {e£)  =  —  [s  +  2.73)  =  —  24.57  tons. 

Moving  load  on  BC  and  DE. 
7?i  =  19-33  tons  ;  hence,  for  the  shear  in  Ef: 

-y  ^  -  /?i  +  (3  +  2  X  7.73  +  2  X  19.5)  =  38.13  tons. 

Hence,  {Ef)  =-s  x  sec  a  =  +49.19  tons. 

Also,  {fF)  =  —  (i-  +  2.73)  =  —  40.86  tons. 

Moving  load  on  BC  and  DP. 
Rx  =  26.40  tons ;  hence,  for  the  shear  in  Fg: 

^=-^i  +  (3  +  3x  773  +  3  X  i9-5)  =  58.29  tons. 

Hence,  (Fg)  =  s  x  sec  a  =  +  75.19  tons. 

Also,  (^6^)  =  —  (s  +  2.73)  =  —  61.02  tons. 

Moving  load  on  BC  and  DG. 
Ri  =  30.14  tons;  hence,  for  the  shear  in  G/i : 

s=—  Ri  +  (t,+4x  7.73  +  4  X  19.5)  =  81.78  tons. 

Hence,  (G/i)  =  s  x  sec  a  =  +  105.5  tons. 

Also,  (/^^)  =  —  (s  +  2.73)  =  —  84.51  tons. 

Moving  load  over  BC  and  AB. 
Rx  =  31.22  tons  ;  hence,  for  the  shear  in  Hd : 

s=  -Rx  +  {s  +  S  ^  7-73  +  5  X  19-5)  =  107.93  tons. 


jq6  ,  "  SWING  BRIDGES. 

Hence,         {Hb)  =  s  x  sec  ^  =  +  117.64  tons. 

The  same  panel  weights  have  been  taken  for  //and  H'  as 
for  D,  E,  F,  etc.,  though,  strictly  speaking,  they  would  be  a 
little  smaller.  At  d,  however,  the  fixed  weight  will  be  taken 
as  2.73  X  7-^13  =147  tons. 

Hence,      (dB)  =  —  (2  x  107.93  +  1.47)  =  —217.33  tons. 

Thus  the  web  stresses,  with  the  exceptions  noticed,  are 
completed. 

It  has  been  shown  that  the  greatest  compression  in  the  up- 
per chord  and  tension  in  the  lower  will  exist  when  the  moving 
load  covers  the  whole  of  one  arm,  as  AB,  for  which  condition 
of  loading,  as  has  already  been  seen : 

i?i  =  44.69  tons. 

Now,  /?!  —  (3  +  7.73  +  19.5)  =  14.46  tons  is  that  part  of  the 
total  panel  load  (fixed  and  moving)  at  E,  which  may  be  con- 
sidered as  passing  directly  to  A  ;  while  (19.5  +  7.73)  —  14.46 
=  12.77  tons  is  the  remainder,  which  maybe  taken  as  passing 
directly  to  B. 

The  following  values  will  now  be  needed : 

14.46  X  tan  a  ~  1 1-75  tons. 
12.77  ^  tana—  10.37     " 
(7.73  +  19.5)  X  tan  a  =  22.12     " 

The  chord  stresses  then  follow : 

{AD)  =  {DE)  =  {R,  -  3)  X  fan  ^  =  +  33.87  tons. 

{de)  =  {ef)  =  —  (2  X  14.46  +  27.23)  X  tan  a=  -  45.62  tons. 

ifg)  =  {</)  +  ^2-77  X  ^an  a  =  —  SS-^S  tons. 

{EF)=-{/g)=  +35.25  tons. 

(£-k)  =  [fg)  +  27.23  X  tana  +  12.77  x  tan  a  =  —  2.76  tons. 

{EG)  =  -  {gh)  =  +  2.76  tons. 


ENDS  SIMPLY  RESTING   ON  SUPPORTS. 


197 


(/!^)  will  evidently  be  tension,  and  {GH)  compression;  no 
other  stresses,  therefore,  are  needed. 

The  following  stresses,  by  moments,  serve  as  checks: 

(de)  =  (./)  =  -  W  -  3)  X  26 -27.23x1 3  ^  _       g3  j„„3. 

^  16 


,    ,,  (^1  —  3)  X  52  -  4  X  27.23  X  IQ.5 

{^gh)  =  —  ^^ ^ ~ — —^ — ^ ^-^  =  —  2.75  tons. 

The  chord  stresses  due  to  the  fixed  load  alone  are  the  fol- 
lowing: 

(AD)  =  —  (de)  =  —  J  X  tan  a  =  -    2.44  tons. 

{D£)  =  -  (ef)  =  (AD)  -  (3  +  7.73)     X  tana=  -  1 1.16    " 

(EF)  =  -  Ifg)  =  {DE)  —  (3  +  2  X  7.73)  x/«w«=  -  26.16    " 

(EG)  =  -  (g/i)  =  {EF)  -  (3  +  3  X  773)  xtana=  -  47.44  " 
{GH)  =  -  {bh)  =  {EG)  -  (3  +4  X  773)  xtana=  -  75.00  " 
{BH)  =  {GH)  -  (3  +  5  X  773)  x  tan  j3  =  -  93.22    " 

As  a  check : 

(BH)  =  -  3  X  72  +  5^x  773  x  33  ^  _  ^^,,  ^ons. 

The  tension   in  the  upper  chord  and    compression  in  the 
lower,  due  to  the  moving  load  only,  still  remain  to  be  found. 

Moving  load  over  AB  and  BC. 
7?i  =  31.22  tons.     Moments  about  b  give: 

{BHY  =  ^1x72-5x19.5  X33  ^  _  60.6  tons. 
16 


Moving  load  over  BC  and  A  G. 
R^  —  30.14  tons.     Moments  about  //  give  : 

/^^/^'  /7L\>         V^i    X    65    —  4    X    19.5    X    32.5 

{GH)  =  -  {lib)'  =  -1 5 — ^^-^^ ^-^  =  -  35-99  tons. 


qg^  SH'/NG  BRIDGES. 

Moving  load  over  BC  and  AF. 
Ri  =  26.40  tons.     Moments  about  g  give  : 

/Z7^\'            /    )\'       ^1  X  52  —  3  X  19.5  X  26  ^  ^ 

i^FG)  =  -  {gh)   =        -^ -J^     ^>* =  -  9.26  tons. 

Moving  load  over  BC  and  AE. 
R^  =  19.33  tons.     Moments  about /"give: 

,r-r-\'  /    /-    ^,         7^11    X    ^9  —  2    X    I9.5    X     1 9.  ^ 

{EF)  =  -  {fg)'=  -i ^ -^-^-5 ^A  =  _  0.4  tons. 


Moving  load  over  BC  and  at  D. 
R^  —  8.39  tons.     Moments  about  e  give: 

/  r^  r-^  .      r^,  -/?1     X     26  —    I9.5     X     I  3 

{DE)  =  -  (r/)  =     ^- g-^^ ^  =  -  2.21  tons. 

•{EF)'  = —{fg)' =  —2.21— (19.5-8.39)  x/«;z  «'= —11,24  tons. 
{EG)'  =  —  (gli)'  -  —  11.24  —  9-03  —  —  20.27  tons. 
{GH)'  =  —  (//Z')'  =  —  20.27  —  9.03  =  —  29.30  tons. 

Other  chord  stresses,  with  the  different  conditions  of  load- 
ing taken,  are  not  indicated,  as  they  were  found  to  be  less,  for 
the  same  panels,  than  those  that  are  given.  They  might  be 
needed,  however,  in  some  cases. 

A  very  important  result  occurs,  which  has  not  before  been 
noticed,  when  the  moving  load  covers  ^C  and  one  panel  load 
rests  at  A. 

In  such  a  case  Eq.  (i)  gives  : 

M=  -  -,^2(P-s')z=  -  486.00. 
4/ 


And  Eq.  (2) 


/?i  =  --  =  -  6.75  tons. 


Under  the  circumstances  just  named,  therefore,  the  condi- 


ENDS  SIMPLY  RESTING   ON  SUPPORTS. 


199 


tion  of  things  at  A  is  equivalent  to  hanging  a  weight  of  6.75 
tons  at  that  point,  as  the  end  of  the  overhanging  arm  AB. 
With  such  a  weight,  the  following  stresses  result : 

6.75  X  tana  —  5.48  tons. 
{AD)"  =  -  {de)"  =  -  5.48  tons. 
(BE)"  =  -  {efY  =  -  10.97  " 
{EFY  =  -  (/-)"  =  -  16.45  " 
{FCy  =  -(^//)"=  -21.94  " 
{GH)"=  -  {bh)"  =  -  27.42  " 
(BH)"  =  —  (27.42  +  6.75  fan  (i)  =  —  30.37  tons. 

From  these  results  are  to  be  selected  the  greatest  chord 
stresses. 

For  examples : 

Resultant  {GH)  —  —  (35.99  +  75.00)  =  —  1 10.99  tons. 
(^^)  =  -(21.94 +47-44)=  -    69.38     " 

In  short,  precisely  as  the  operation  has  been  done  before. 
The  resultant  web  stresses  caused  by  this  negative  reaction, 
are: 

{Ad)  =  (3  +  6.75)  y.  sec 01  —  ^  12.58  tons. 
{cW)  =  -  (3  +  2.73  +  6.75)  =  —  12.48  tons. 

These  are  the  "  singular  "  stresses  already  mentioned. 
Collecting  and  arranging  the  results,  the  following  resultant 
stresses  are  obtained  : 


{dE)   =  +     23.19  tons. 

{Ad)=  +     12.58     " 

{Ad)  =  -     53.78  tons 

{dD)  =  +    24.50     " 

{dD)   =  -     12.48     " 

{De)    =  +    28.17     " 

{eE)   =  -    24.57     " 

{Ef)  =  +    49-19     " 

{fF)  =  -    40.86     " 

{Fg)  =  +     75.19     " 

{gG)  =  —    61.02     " 

{Gh)  =  +  105.50     " 

(/.//)=  -    84.51     " 

{Hb)  =  +  117.64    *" 

{bB)   =  -  217.33     " 

{AD)  =  -      7-92 

tons;    +  33.87  tons. 

{DE)  =  -    22.13 

"  ;     +  33-87     " 

200  SWING  BRIDGES. 


{EF)  =  -    42.61  \ 

tons ; 

+  35.25  tons, 

{FG)   =  -  69.38 

"  ) 

+  2.76  " 

{GH)  =  —  1 10.99 

(/?//)= -153-82 

(de)     =  +      7.92 

—  45.62  tons. 

(</■)  =  +    22.13 

-  45.62  - 

{/£■)    =  +  42.61 

-  35-25  " 

i^/i)     =  +    69.38 

-  2.76  " 

(/id)     =  +  110.99 

The  same  stresses  exist,  of  course,  for  corresponding  mem- 
bers in  the  arm  BC. 

It  is  thus  seen  that  the  portions  dk,  d'h' ,  A  G,  and  CG' ,  of 
the  chords  must  be  counterbraced.  Ad,  Cd\  dD,  and  d'D , 
only,  of  the  web  members  need  the  same  treatment. 

It  may  happen  that,  with  a  moving  panel  load  at  Z>,  the 
reaction  at  A  will  be  negative,  in  the  search  for  main  web 
stresses.  In  such  a  case  the  method  of  operation  is  simply 
an  extension  of  that  used  in  the  example.  If  the  numerical 
value  of  this  negative  reaction  is  equal  to,  or  less  than,  a 
moving  panel  load  (which  may  rest  at  A),  a  weight  equal  to 
this  reaction  is  to  be  taken  as  hung  from  A,  and  the  panel 
load  (moving)  at  D  is  to  be  taken  as  hung  from  that  point, 
while  the  arm  AB  is  to  be  considered  as  an  overhanging  one. 
If  the  reaction,  however,  is  greater  than  a  moving  panel  load, 
then  two  such  loads  are  to  be  taken  as  hanging  from  A  and 
D  with  the  overhanging  condition  of  the  arm, 

A  whole  panel  moving  load  is  taken  at  A  for  prudential 
reasons.  If  the  load  were  of  uniform  density,  then  a  half 
panel  moving  load  would  be  taken  at  A. 

Negative  reactions  by  the  formula  for  any  number  of  mov- 
ing panel  loads  near  the  end  are  to  be  treated  in  exactly  the 
same  way  ;  for  it  is  to  be  remembered  that  negative  reactions 
in  an  actual  truss,  in  this  case,  cannot  exist. 

It  may  be  urged  that  the  case  of  partial  continuity,  taken 
in  the  preceding  Article,  should  be  treated  according  to  the 
principles  developed  in  this,  by  taking  the  middle  span  equal 
to  zero. 


ENDS  SIMPLY  RESTING  ON  SUPPORTS. 


20 1 


Making  such  an  assumption,  however,  would  be  a  depart- 
ure from  the  real  state  of  the  truss.  The  safe  way  would  be 
to  determine  the  greatest  stresses  by  both  methods,  and  select 
the  greatest  of  the  two  sets  of  results. 

Differences  would  be  found  only  in  the  upper  chord  ten- 
sion, lower  chord  compression,  and  main  web  stresses. 

In  the  present  case,  if  there  are  two  or  more  systems  of 
triangulation,  each  is  to  be  treated  precisely  as  the  example 
has  been. 

This  case  really  includes  that  of  a  centre-bearing  turn-table 
with  two  points  of  support  at  the  centre,  as  shown  in  the 


h.    Ti- 


figure.  HH'  is  free  to  "  rock  "  on  the  central  point  B,  and  as 
the  motion  is  always  very  small,  BH  (horizontal  distance)  is 
essentially  equal  to  BH'  (also  horizontal)  at  all  times.  From 
this  it  results  that  the  reaction  at  H  will  always  be  equal  to 
that  at  //',  consequently  the  diagonals  Hh  and  H'h  must  be 
introduced. 

Now  as  HH'  is  really  a  part  of  the  truss,  attached  to  and 
moving  with  it,  the  whole  bridge,  AC,  is  simply  a  continuous 
truss  of  two  spans  supported  on  the  fixed  point  B.  All  the 
conclusions  and  formulae,  therefore,  of  this  Article,  apply  to 
it  directly.  R^  will  be  the  reaction  at  B,  and  M  will  be  the 
moment  over  the  same  point. 

According  to  the  principles  established,  HJi  will  receive  its 
greatest  stress  when  AB,  only,  carries  moving  load.  Since 
the  pressure  on  H  is  always  equal  to  that  on  H'  also  to  a 
half  of  the  reaction  at  B,  there  results  : 


{Hh)  = 


R'9  X  sec  hHh 


in  which  i^'g  is  the  reaction  at  B  due  to  the  moving  load 


on 


202  SWING  BRIDGES. 

AB  only,  considered  as  a  simple  truss.  The  greatest  stress 
in  H'h  is,  of  course,  equal  to  {Hh). 

The  greatest  stress  in  Hh  (equal  to  [H'h'))  is  found,  as 
before,  by  putting  the  moving  load  on  BC  and  AG. 

No  locomotive  excess  has  been  taken,  but  precisely  the 
same  conditions  of  loading  hold  whether  such  excess  is  taken 
or  not. 

It  will  only  be  necessary  to  remember  that  the  locomotive 
may  be  at  either  end  of  the  train,  and  that  the  greatest  results 
arising  from  the  two  positions  are  to  be  selected. 


CHAPTER    V. 

SWING   BRIDGES.      ENDS    LATCHED   TO    SUPPORTS. 
Art.  39. — General  Considerations. 

It  has  already  been  stated  that  the  object  of  fitting  the 
ends  of  a  swing  bridge  with  a  latching  apparatus  is  to  enable 
those  ends  to  resist  a  negative  reaction,  or  in  other  words,  to 
prevent  their  rising  from  the  points  of  support.  All  "ham- 
mering "  of  the  ends  will  thus  be  prevented. 

It  has  further  been  shown  in  the  preceding  Chapter  that  if 
there  are  always  two  points  of  support  at  the  center,  for  each 
system  of  triangulation,  the  ends  will  never  tend  to  rise.  It 
was  also  observed  in  the  preceding  Article  that  with  a  pivot, 
or  centre-bearing  turn-table,  the  bridge  always  presents  the 
case  of  continuity  with  two  spans  only,  whatever  may  be  the 
number  of  apparent  points  of  support  at  the  centre. 

In  this  chapter,  then,  it  will  only  be  necessary  to  consider 
the  one  case  of  continuity  with  a  single  point  of  support 
between  the  extremities  of  the  bridge. 


Art.  40.— Ends  Latched  Down— One  Point  of  Support  Between  Extremi- 
ties  of  Bridge — Example. 

The  general  formulae  required  in  this  case  are  Eqs.  (i),  (2), 
(3),  and  (4),  of  Article  38,  and  they  are  here  reproduced. 


M 


ie,  =  i|i^p(/-.i  +  7i/| (2). 

203 


204  SWING  BRIDGES. 

/?2-^|i/'^  +  i^/^^-2J/| (3). 

i?3  =  y|i^/^(/-^i  +  ^/| (4). 

These  involve  the  condition  1^  =  1^  —  /,  which  will  appear  in 
the  example. 

If  this  condition  does  not  exist  in  any  case,  the  formulae  to 
be  used  are  Eqs.  (lo),  (ii),  (12),  and  (13)  of  Article  35,  but 
they  are  to  be  used  in  precisely  the  same  manner  as  will  be 
Eqs.  (I),  (2),  (3),  ajid  (4). 

This  case  was  essentially  treated  in  the  preceding  Article, 
insomuch  that  with  ordinary  moving  loads  precisely  the  same 
conditions  of  loading,  for  the  greatest  stresses,  are  required  in 
the  two  cases.  The  results  themselves,  however,  will  be  dif- 
ferent for  the  upper  chord  compression,  lower  chord  tension, 
and  counter  stresses. 

It  will  probably  be  as  expeditious  and  labor  saving,  never- 
theless to  find  the  reactions  and  chord  stresses  due  to  each 
moving  panel  load,  and  then  combine  the  results  thus  found 
with  those  due  to  the  fixed  load  alone,  in  the  usual  manner. 
Such  is  the  method  to  be  used  in  the  example. 

Since  Eq.  (i)   shows    that  M  is  always  negative,   Eq.  (2) 

1 
shows  that  with  a  given  value  of  2P{1  —  z),  R^  will  have  its 

greatest  positive  value  when  no  moving  load  is  upon  the  span 

/g.     The  expression  for  the  shear  in  any  counter: 

s=  R,-n'u'-n{lV+  IV'); 

(in  which  n'  is  the  number  of  moving  loads  between  R^  and 
the  counter,  and  n  the  number  of  fixed  loads  similarly  lo- 
cated), will  have  its  greatest  value  for  ;/'  =  o.  Hence, /<?r  t/ie 
greatest  stress  in  any  inclined  counter^  the  moving  load  must  ex- 
tend from  the  centre  to  the  foot  of  the  counter  in  question.  This 
is  precisely  the  condition  used  previously. 

As  usual,  the  stress  in  the  vertical  which  cuts  the  upper 


ENDS  LATCHED   DOWN: 


20  q 


extremity  of  the  counter  must  be  found,  for  it  may  be  the 
greatest  in  that  member. 

Precisely  the  same  reasoning  used  previously  shows  that 
the  greatest  stress  in  a7iy  main  tvcb  member  {inclined)  exists 
when  the  moving  load  eovers  the  whole  of  one  span,  and  that 
portion  of  the  other  included  between  the  free  end  and  the  foot  of 
the  member  considered. 

The  stress  in  the  vertical  which  cuts  the  upper  extremity 
of  the  inclined  web  member  is  to  be  found  with  the  same 
condition  of  loading  ;  it  will  usually  be  the  greatest  possible. 

The  truss  to  be  taken  for  an  example,  and  all  the  data,  are 
exactly  the  same  as  those  used  in  Article  38.  The  figure  and 
the  data  are  reproduced  below : 


Total  length  =  AC  ^  2I  ^  2AB  =  2BC  =  144  feet. 
Uniform  depth  =  dD  =  bB  =  \6  feet. 
Panel  length  =  AD=  DE  =  etc.  =  13  feet. 

BH=BH'  =^7  feet. 
Total  fixed  weight  per  foot  =  1200  pounds  (nearly). 
Upper  chord  panel  fixed  weight  =  JV=  2.73  tons. 
Lower  chord  panel  fixed  weight  =  W  =  5.00  tons. 
Uniform  panel  moving  load  =  w=  19.50  tons. 

The  inclined  w^eb  members,  except  the  end  posts,  are  for 
tension,  and  the  verticals  for  compression. 

The  moving  load  traverses  the  lower  chord.  The  weight 
of  the  floor  system  is  taken  at  about  350  pounds  per  foot  of 
track.  The  fixed  weight  at  A  will  be  taken  at  three  (3)  tons, 
and  that  at  b  at  1.47  tons.  Full  panel  loads  of  both  kinds 
will  be  taken  at  //"and  //'. 

For  a  single  moving  panel  load  on  the  arm  AB: 


M=-^,iP 


^): 


(5). 


2o6 


SWING  BRIDGES. 


^1  = 


i|z^(/-..)+J/| 


R.^'l 


(6). 


(7). 


The  distance  z  is  to  be  measured  from  A  ox  C  according  as 
the  arm  AB  ox  CB  is  considered. 

The  trigonometrical  quantities  used  in  this  example  are 
the  same  as  those  employed  in  the  preceding  Article.  They 
are  the  following: 

tan  AdD  =  tan  a  =  0.8125 
sec      "      =^  sec  a=i.2g 
tan  HbB  =  tan  ft  =  0.4375 
sec      "      =  sec  (d  =  \  .09 

The  following  quantities  are  also  taken  from  the  example 
in  the  preceding  article : 


z=  13 
^  =  26 

.     .     /-z  =  S9     .     .     {/' 

—  z^)z=    65195.00 
_  s^)z=  117208.00 

^=39 

^=52 

z  =  6$ 

.     .     /-.^  =  33     •     .     (Z^- 
.     .     /-.=    7     •     •     i^'- 

-z^)z=  142857.00 

-  z^)z=  128960.00 

-  z^)z  =    62335.00 

4/^^20736=°-^^4823. 

0.00004823  X  19.5=0.00094. 

w       19.5 

—  =^^  =  0.271. 
/       72 

By  using 

these  quantities  in  Eqs.  (5) 

(6),  and  (7): 

w  dX  D     . 

.     Ri=  +  15.14  tons     .     . 

R^=  —  0.851  tons. 

w  ''  E     . 
w  ''  F     . 
w  ''  G     . 
w  ''  H   . 

.     Ri=  +  10.94      "       .     . 
.     R,=  +    7.07      "       .     . 
.     R^=  +     3.74      "       .     . 
.     ^,  =  +     1.09      " 

A=-i-53 
i?3=-i.87 
/?3=-i.68 
y?3=-o.8i 

:^R,  =  +  37.98 

:^i?3= -6.741      " 

ENDS  LATCHED  DOWN. 


207 


The  greatest  negative  reaction  at  the  extremity  of  one  arm 
will  exist  when  the  whole  of  the  other  is  covered  by  the  mov- 
ing load,  and  its  value  is  seen  to  be  —  6,741  tons.  The  resist- 
ance of  the  latching  apparatus  must  be  sufficient  to  oppose 
this  with  a  proper  safety  factor. 

Under  the  same  circumstances,  with  ends  not  latched  down, 
it  was  found  that  the  reaction  dXA  was  44.69  tons  (see  Article 
38);  but  37.98+67.41=44.721  tons,  which  is  essentially 
equal  to  44.69  tons,  as  it  should  be. 

Counter  Stresses. 
dE  is  the  only  counter  needed,  since  with  moving  loads  at 
E^  F,  G,  H,  the  reaction  2X  A  is: 

R^  =  10.94  +  7.07  +  3.74  +  1.09  =  +  22.84  tons  ; 

consequently  2P  =  o  at  E. 

The  shear,  or  vertical  component  of  the  stress,  in  dE  is: 

j=  22.84  —  (3  +  -73  +  5-00)  =  12. 1 1  tons. 

'  Hence,         {dE )  =  s  >^  sec  a  =  +  15.62  tons. 

With  the  moving  load  covering  DH,  dD  acting  as  a  counter 
will  sustain  a  tensile  stress  equal  to 

[dD)  =  +  (19.5  +  5.00)  =  4-  24.5  tons. 

With  the  same  condition  of  loading.  Ad  receives  its  great- 
est compressive  stress  : 

{Ad)  =  —Ri  X  sec  a  =  —  37.98  sec  a  =  —  48.99  tons. 

Main  Web  Stresses. 

The  main  web  stresses  are  found  precisely  as  in  Article  38, 
and  there  is  no  need  of  repeating  the  operation  here. 

The  values  of  the  stresses  will  be  reproduced  in  the  proper 
place. 

Chord  Stresses. 
The  chord  stresses  due  to  the  fixed  load  alone  are  the  same 
as  those   determined   on  page  195  of  Article  38.     They  will 


2o8  SWING  BRIDGES. 

not  De  reproduced,  but  references  will  be  made  to  them  as 
they  are. 

Those  caused  by  the  moving  load  alone  will  be  determmed 
by  placing  a  panel  load  at  each  panel  point  successively, 
and  finding  all  the  chord  stress  in  both  arms  due  to  it,  then 
tabulating  the  results,  and,  finally,  combining  them  in  the 
manner  already  shown  in  several  instances. 

The  counter  dE  will  be  supposed  to  come  into  action  for 
the  weights  E,  F,  G,  and  H. 

The  panel  loads  at  F,  G,  and  H  will  cause  apparent  com- 
pression in  some,  or  all,  of  the  inclined  members  Ef,  Eg,  and 
Gh.  The  resultant  action  of  fixed  and  moving  loads  in  those 
members,  however,  will  in  all  cases  be  tension. 

The  detailed  expressions  for  the  chord  stresses  due  to  one 
moving  panel  load  only  will  be  given,  as  all  the  others  are 
like  it.     For  this  purpose  take  w  ^X  E. 

7?i  =  +  7.07  tons  ;  R^=  —  \.2>y  tons, 

w  —  Ry^—  19.5  —  7.07  =  12.43  tons. 
{AD)  —      {DE)  —  +  Ry  tan  «  =  +  5.74  tons. 
{de)     =       {ef)    =  —  2  X  i?i  X  tan  a  —  —  11.49  tons. 
{EE)  =  -  (fg)    =  +  1 1.49  +  5.74  =  +  17.23  tons. 
{EG)  —  —  {gh)    =  17-23  —  12.43  X  tana  =  +  7.13  tons. 
{GH)=  —  {hb)    =    7-13  —  12.43  X  i<^^  a  =  —  2.97  tons. 
{HE)  =  —  2.97  —  12.43  X  t^n  ft  =  —  8.41  tons. 

{CD')   =  —  {d'e')  —         R3  X  tana—  —  1.52  tons. 
{DE')  =  -  {e'f)   =  2  X  "  "       =  -  3.04      " 

{E'F)  =  -  yfg)  =  3  X  "  "      =  -  4-56     " 

{F'G')  =  -  {gh)  =  4  X  "  "      =  -  6.08     " 

{G'H')  =  -  {kb)   =  5  X  "  ''      =  -  7.60     " 

{H'B)  =  -  7.60  -  RsX  tanft  =  -  8.42      " 

The  following  checks  by  moments  should  be  observed. 
Moments  about  b  give  : 

/zj-zjN        Ri  X  72  ~  19.5  X  33  o        . 

{HB)   —  — ^ '- ^  ^ ^  =  —  8.40  tons  ; 


{H'B)  =  ^^4^ 
^  16 


=  —  8.415  tons. 


ENDS  LATCHED  DOWN.  209 

Moments  about  f  give  : 

{EF)  —     ^      ^^  =  +  17.23  tons. 

The    following   tables   are  found    by   following  the   same 
operation  for  all  the  weights. 


{de) 

{ef) 

(/^) 

igk) 

{hb) 

w  ^t  D 

—  12.3 

—    8.76 

—  5-22 

-  1.68 

+  1.86 

"  "  E 

-  17-78 

-  17-78 

-  10.82 

-  3-87 

+  3-09 

"  "  F 

-  11-49 

-  11-49 

-  17-23 

-  7-13 

+  2.97 

"  "  G 

-  6.08 

-  6.08 

—  9.12 

—  12.16 

+  0.65 

"  "  JI 

-  1-77 

-  1-77 

-  2.66 

-  3-55 

-4.42 

{AD) 

{DE) 

{EF) 

{FG) 

{GH) 

{HB) 

W  3l.\.  D 

+  12.3 

+  8.76 

+   5-22 

+    1. 68 

-  1.86 

-3-77 

"  "  E 

+  8.89 

+  8.89 

+  10.82 

+  3-87 

-  3-09 

-6.94 

"   "  F 

+  5-74 

+  5-74 

+  17-23 

+  7-13 

-  2.97 

—  8.41 

"  "  G 

+  3-04 

+  3-04 

+  9.12 

+  12.16 

—  0.65 

-  7-54 

"  "  H 

+  0.89 

-f  0.89 

4-  2.66 

+  3-55 

+  4  42 

-  3-63 

{de) 

(e'f) 

i/Y) 

(g'A') 

{A'b) 

w  at  D 

+  0.69 

+  1-38 

4-2.07 

4-2.76 

+  3-45 

"  "  E 

+  1.24 

+  2.49 

+  3-73 

+  4-97 

+  6.22 

"    "   F 

4-1.52 

+  3-04 

+  4-56 

-1-  6.08 

4-7.60 

"    "  G 

4-  1-37 

4-2.70 

4-  4.10 

4-  5-46 

-f  6.83 

"    "  H 

4-  0.66 

4-1.32 

4- 1.97 

4-  2.63 

4-  3-29 

14 


210 


SWING  BRIDGES. 


(CD) 

(DE) 

(£F') 

(FG) 

iC/l) 

{HB) 

7f/al  D 

—  0.69 

-1.38 

—  2.07 

—  2.76 

-  3-45 

—  3-82 

"  "  E 

-1.24 

-  2  49 

-  3-73 

-  4-97 

—  6.22 

-6.88 

"    "   /r 

-1.52   ! 

-  304 

-4-56 

-  6.08 

—  7.60 

—  8.42 

'■    "    G 

-1-37    i 

-2.73 

—  4.10 

-  5-46 

-  6.83 

-756 

"   "  J/ 

-0.66 

—  1.32 

-  1-97 

—  2.63 

-  329 

-  364 

Using  the  main  web  stresses  and  the  fixed  weight  chord 
stresses  found  in  Article  38,  the  following  greatest  stresses  at 
once  result : 


idE) 

=  +     15.62  tons. 

{Ad) 

=  +     12.58 

(< 

{Ad)  =  —    48.99  tons 

{dD) 

=  +    24.50 

It 

{dD)  =  -    12.48      " 

{De) 

=  +    28.17 

u 

{eE)  =  -    24.57      " 

im 

=  +    49-19 

<< 

i/E)  =  -    40.86      " 

{Fg) 

=  +  . 75-19 

<< 

{gG)=  -    61.02      " 

{Gh) 

=  +  105.50 

<< 

{/iB)=-    84.51      " 

{Hb) 

=  +  117.64 

(( 

{bB)  =  -  217.33      " 

{AD)  =  - 

7.92 

ton. 

5 ;          4-  28.42  tons. 

{DE)  =  - 

22.12 

<< 

;          +  16.16     " 

{EF)  =  - 

42.59 

(( 

;          +  18.89     " 

{EG)   =- 

69-34 

(( 

{GH)=- 

I  10.96 

li 

{BH)=- 

153.83 

a 

{de)     =  + 

7.92 

a 

;         —  46.98  tons. 

(</)     =  + 

22.12 

i( 

;       -  34-72    " 

ifg)    =  + 

42.59 

a 

;         -  18.89     " 

{g^^)    =  + 

69-34 

a 

{/lb)    =  + 

I  10.96 

(( 

The  web  members  dD,  d'D',  Ad,  Cd\  and  the  portions  AF^ 
CF',  dg,  d'g'  of  the  chords  need  counterbracing. 


ENDS  LA  TCHED   DO  WN.  2 1  r 

The  chord  stress  {GH)  —  —  110.96  tons  requires  the  mov- 
ing load  to  cover  BC  and  AG. 

With  the  moving  load  on  AB  only,  there  is  some  ambi- 
guity in  the  stresses  {dc),  compression,  and  {DE)  tension. 

In  such  a  case  the  reaction  R^  is  37.98  tons,  and  the  web 
member  De  may  be  neglected.  Under  such  an  assumption, 
by  taking  moments  about  d  and  E  successively,  there  will 
result : 

{AD)  =  {DE)  =  ^^'~^^  ""  ^^  =  +  28.42  tons ; 

(de)  =  {e/)  =  -  ^-i ^^ F6~~"^ =  ~  ^472  tons. 

This  ambiguity  cannot  be  avoided  if  both  the  web  mem- 
bers dE  and  De  exist.  It  might  also  have  been  noticed  in 
the  case  last  treated. 

It  has  already  been  noticed  that  the  downward  reaction  of 
6.74  tons  must  be  resisted  by  the  latching  apparatus. 

If  there  are  two  or  more  systems  of  triangulation,  the  pre- 
ceding principles  hold  true  for  each.  Also,  if  there  is  loco- 
motive excess,  precisely  the  same  methods  are  to  be  em- 
ployed. 

The  observations  which  were  made  at  the  end  of  Article  38 
on  a  pivot  or  centre-bearing  turn-table,  over  which  there  are 
two  points  of  support  for  the  truss,  apply,  exactly  as  they 
stand,  to  this  case.  The  value  of  R'2  must,  however,  be  found 
by  Eq.  (3)  of  this  Article. 


CHAPTER    VI. 

SWING   BRIDGES   ENDS   LIFTED. 
Art.  41. — General  Considerations. 

In  the  preceding  chapter  there  was  noticed,  in  detail,  the 
method  of  prevention  of  "  hammering,"  by  latching  down  the 
ends  of  a  swing  bridge  of  two  spans.  It  was  also  there 
noticed  that  the  necessity  of  such  an  arrangement  could  only 
exist  in  the  case  of  continuity  with  two  spans.  For  precisely 
the  same  reasons  given  in  connection  with  that  case,  the  neces- 
sity of  lifted  ends  can  exist  in  the  event  of  continuity  with  twa 
spans  only. 

It  is  plain  that  if  the  ends  of  a  swing  bridge  are  pressed 
upward  by  forces  exceeding  the  greatest  negative  reactions 
determined  for  latched  ends  by  the  formulae  of  the  last  chap- 
ter, there  can  be  no  hammering,  for  the  ends  can  never  leave 
their  seats  or  supports. 

By  a  proper  device,  then,  the  ends  should  be  pressed  up- 
ward by  forces  at  least  equal  to  the  negative  reactions  deter- 
mined for  latched  ends. 

In  order  to  provide  for  any  contingency,  however,  which 
may  arise,  the  upward  force  should  somewhat  exceed  such  a 
value. 


Art.  42.— Ends  Lifted— One  Point  of  Support  Between  Extremities- 
Example. 

The  figure  represents  the  truss  to  be  taken  as  an  example. 
The  span,  depth  of  truss,  and  panel  lengths,  excepting  hh'y 


J)        H       \F         G- 


ENDS  LIFTED. 


2\y. 


are  the  same  as  those  taken  in  the  two  preceding  cases ;  the 
loading  is  also  the  same. 

The  following  are  the  data  to  be  used  : 

AC=2(Ah+  -—)=  144  feet. 

Uniform  depth  of  truss  =    16     " 

Panel  length  =13     " 

hh'  =    14     " 

Uniform  fixed  upper  chord  panel  load  =  IV  =    5.00  tons. 
"      lower       "         "         "     =  W=    2.73     " 
"  moving  "         "     =  w    =  19.50     " 

Moving  load  for  unit  of  length  =    1.50     ** 

The  truss  is  a  deck  one,  as  the  moving  load  passes  along 
AC;  and  as  the  figure  shows,  there  are  two  systems  of  tri- 
angulation.  It  will  be  assumed,  though  not  strictly  true,  that 
the  same  panel  loads  are  found  at  /i  and  k'  as  at  the  other 
panel  points. 

Let  the  inclination  of  G/i  to  a  vertical  line  be  denoted  by  a. 


Bh 

(I 

<(               (< 

A 

Bg 

<( 

«              « 

<^. 

Then  tan  a  =  0.8125  ; 

sec  a  =  1.29; 

-        "   ^  =  0.4375; 

"  /?=  1.09; 

"    d==  1.25; 

''6=  1.6. 

Each  system  of  triangulation  is  to  be  treated  as  an  inde- 
pendent truss.  The  fixed  weights  at  D  and  A  will  be  taken 
as  belonging  to  the  system  ADeF,  etc.,  while  that  at  ^  will  be 
assumed  to  belong  to  the  other  system.  Similar  observations 
apply  to  the  other  arm.  As  in  the  preceding  case,  a  fixed 
load  of^three  (3)  tons  will  be  taken  at  A  or  C. 

The  stresses  in  a  swing  bridge  with  ends  lifted  may  be 
considered  as  composed  of  the  stresses  in  two  other  trusses, 


2j.  SWING   BRIDGES. 

one  with  ends  latched  down  and  subjected  to  the  same  loads 
(both  fixed  and  moving),  and  the  other  subjected  to  the 
action  of  the  upward  pressures  only,  at  the  ends ;  the  dif- 
ferent trusses  being  supposed  of  the  same  form  and  dimen- 
sions in  all  their  parts. 

From  this,  it  at  once  follows  that  the  positions  of  the  mov. 
ing  load  for  the  greatest  stresses  {when  the  ends  are  lifted) 
are  exactly  the  same  as  those  determined  in  the  preceding 
chapter. 

For  the  stresses  in  the  counters,  then,  or  for  those  in  the  mem- 
bers which  slope  downward  from  the  upper  chord  and  toward 
the  ends,  the  moving  load  must  extend  from  the  centre  to  the 
upper  extremities  of  such  members. 

These  stresses  will  be  compressive,  and  the  member  in 
which  such  stress  is  first  found  is  to  be  determined  in  the 
manner  already  shown. 

In  order  to  find  the  greatest  compressive  stress  in  any  web 
member,  in  one  arm,  sloping  downward  from  the  upper  chord, 
and  toward  the  centre,  the  moving  load  must  extend  from  the 
end  of  that  arin  to  its  upper  extrem.it y,  and  at  the  same  time 
cover  the  whole  of  the  other  arm. 

These  conditions  of  loading  are  to  be  taken  while  the  ends 
are  lifted,  but  it  will  also  be  necessary  to  find  the  web  stresses 
for  the  open  draw  in  the  vicinity  of  the  end,  as  some  of  these 
will  be  the  greatest  stresses  in  the  web  members  there  located. 

It  is  to  be  borne  in  mind  that  any  two  web  members  which 
intersect  in  that  chord  which  does  not  carry  the  moving  load, 
take  their  greatest  stresses  together. 

Although  positions  of  moving  load  for  the  greatest  chord 
stresses  may  be  assigned,  it  will  probably  be  the  shortest  and 
most  labor-saving  method  to  find  the  chord  stresses  due  to 
the  fixed  load  and  upward  pressure  together,  then  find  those 
due  to  each  moving  panel  load  alone,  and  combine  the  results. 
This  method  will  be  used. 

The  example  will  now  be  treated. 

The  following  quantities  are  determined  on  the  supp£)sition 
that  the  ends  are  latched  down,  by  Eqs.  (i),  (2),  and  (4)  of 
Article  40. 


EA'DS  LIFTED. 

System  ADdEf,  etc. 

2=13  feet  . 

,   M=  —    61.28  .  .  >?i  =  +  15.14  . 

.  Rs  = 

—  0.851  tons. 

2  =  39    "     ■ 

.   .']/=  -  134-29  .  .  i?,  =  +    7.07  . 

.  ^3  = 

-1.87       " 

2  =  65    '■     . 

.  M-  -    58.59  .  .  i?i  =  +     1.09  . 

.  ^.  = 

—  0.81 

21 


3-531 


System  ADeFg,  etc. 


=  26  feet  . 

.  M  —  —  110.18  . 

.  i?i  =  +  10.94  . 

-  R^  =  —  1.53  tons. 

=  52    "     . 

.  M  =  —  121.22  . 

.  ^1  =  +    3-74  - 

.  i?^  =  -  1.68     " 

3.21 


Each  of  these  results,  it  is  to  be  obsen^ed,  is  for  a  single 
panel  moving  load  placed  at  the  panel  point  denoted  by  the 
value  of  z.     They  have  been  used  in  the  two  preceding  cases. 

The  chord  stresses  due  to  each  panel  moving  load  alone 
will  first  be  found.  As  these  are  all  found  by  exactly  the 
same  method,  the  detailed  expressions  for  two  only  (one  in 
each  system)  will  be  given. 

System  ADdEf,  etc. 
Panel  moving  load  at/": 

^  =  39  feet ;         Rx—  -^r  7.07  tons  ;  i^g  =  —     1.87  tons. 

{Ad)    —  —  R^  tan  a  =  —     5.74     " 

{df)      =  —  2  Ri  tan  a  =  —  1 1.49     " 

{fh)  =  —  11.49  +  (12.43 — 7.07)  tan  n  —  —  7.20  " 
{hh')  =  —  7.20+  i2.43(/(2«a' + /<:^;^/S)=  +  8.30  " 
{DE)  =  -{Ad) 
{EG)  =  5.74  +  2  i?i  tan  a 
{GB)  =  17.23  —  2  X  12.43  ^  tana 
(Cd)  =  —  R^  tan  a 
{df)  =  -  2  "  " 
{fJi)  =  -  4  "  " 
{hh')  —  6.08  —  ^3  {tan  a  +  tan  ft) 
{DE')  =  -  {Cd') 
{E'G')  =  -f-  3  i'^g  tan  a 
{G'B)  =  +  S  ''      " 


=  + 

5-74 

=  + 

17-23 

=  — 

2.96 

=  + 

1.52 

=  + 

3 -04 

=:    + 

6.08 

=    + 

8.42 

=    — 

1.52 

=    — 

4.56 

=    — 

7.60 

2i6  SWING  BRIDGES. 

As  numerical  checks,  the   moment   method  gives  the  fol- 
lowing results : 

,  „„-        7^1  X  65  —  19.5  X  26  ^  ^ 

ifiB)  =  —^ — ^-^^ =  —  2.96  tons. 


(M')  = ^-^^  =  +8.41  tons. 


(G'B)  =  ^^^  =  -  7.60  tons. 


System  ADeFg,  etc. 
Panel  moving  load  at  e: 

^=26  feet;         7?i  =  +  10.94  tons  ;  R^=  —     1.53  tons, 

(Ae)       =  —  Rx  tan  a  =  —    8.89     " 

[eg)        =  {Ae)  —  2.38  X  taji  a  =  —  10.82     " 
igg')      =  (^<r)  +  ( 19-5  -  ^1)  {tan  a  +  tan  6)  =  +    6.84     " 

{DF)     =  2^1  tan  a  =  +  1778     " 

{FB)      =  (DF)  -  2  (19.5  -  F,)  tana  =  +    3.87     '' 

{Ce)      =  —     7?3  tana  =  +     1.24     " 

{e'g)     =  -  3  -       "  =  +     3.73     " 

igg)     =  (^'<^')  -  ^3  {tan  a  +  tan  6)  =  +    6.88     " 

{DF')  =  2  i?3  Z^;?  a  =  -    2.49     " 

(F'^)    =4"     "  =_    4.97     " 

Moments  give: 

/  Z7D\      /?!  X  52  —  iQ.t;  X  26  o^  . 

{FB)  =  -i — 5 y_5 =  4.  3.86  tons» 

16  ^ 


(/T'^)  =  ^^  =  -  4.97  tons. 


EiVDS  LIFTED. 


2I> 


All  the  results  for  the  two  systems  give  the    four   tables 
below : 


W  ZX  d 

{Ad) 

W) 

ifh) 

{hh) 

{DE) 

{EG) 

{GB) 

—  12.30 

-  8.76 

-  1.68 

+  3.77 

+  12.30 

+    5-22 

-  1.86 

w  zX.  f 

-   5-74 

-11.49 

—   7.20 

+  8.30 

+   5-74 

+  17-23 

—  2.96 

w  3.t  h 

-  0.89 

-  1.77 

+  3.63 

4-  0.89 

+  2.66 

+  4-43 

{Cd) 

{df) 

(/VO 

{hh') 

{DE) 

{EG) 

{G'B) 

w  2X  d 

+  0.69 

+  1.38 

+  2.77 

+  3.83 

—  0.69 

—  2.07 

-'3-46 

W  2X  f 

+  1.52 

+  3-04 

+  6.08 

+  8.42 

-  1-52 

-4.56 

—  7.60 

■w  zX  h 

+  0.66 

+  1.32 

+  2.63 

+  3.64 

-0.66 

+  1-97 

-  329 

{Ae) 

-8.89 
-  3  04 

{eg) 

—  10.82 

—  9.12 

igg) 

{DF) 

{FB) 

+     3-87 
+  12.16 

w  aX  e 
"  "  E 

+  6.84 
+  7-54 

+  17-78 
+    6.08 

{Ce) 

{eg) 

ig'g) 

{DF)          {FB) 

TO  at  ^ 

+  1.24 
+  1-37 

+  3-73 
+  4.10 

+  6.88 
+  756 

-2.49         -4-97 
-2.73         -546 

The  open  draw  stresses  due  to  the  fixed  weight  alone  are 
the  followine : 


3      X  tan  a  =  2.44  tons. 
W  X      "      =4.06    " 

JV  X      "      =  2.22    " 


3      X  tand  =  3.75  tons. 
IV  X       "     =  6.25     " 
W  X      "     =  3.41     " 


2i8  SWING   BRIDGES. 

W  X  /an /3  =  2.  ig  tons. 
W  X      "      =  1. 19     " 

(Ad)  =  +  3  X  tan  a  =  +    2.44  tons. 

(de)     ={Ad)+  JV  tan  a  =  +     6.50    " 

(e/)     =  (de)  +  |2  (3  +  W)  +  W\  tana  =  +  19.88     " 

{fg)    =  (V)  +  ]2  {W  +  W)  +  W\  tan  a         =  +  36.50    " 
ig^i)    =  ifg)  +  (3  +  f'^  +  2  W)  {tan  a  +  /««  6)  +  /F  Z^?/  <? 

=  +70.51  tons. 
{hh)   =  {gh)  +  2  {IV  +VV')  {tan  a  +  /««  /:^)      +  W tan  (i 

=  +  92.02  tons. 
{DE)  =  —  (2x3+  W)  tan  a  =  —    7.10    " 

{EF)  =  (i^-fi")  —  2W  tan  a  —  W  tan  a  =  —  17.44    " 

{FG)  =  {EF)-  2(3+  W^+  lV')tana-  JV'tana=  -  37.10    " 
{GB)  =  {EG)  -  {4IV  +  3IV')  tana  =  -  60.00    " 

As  a  numerical  check: 

{GB)  -{iW  +  2  W)  tan  ^  -  {^  +  2IV  +  2  IV ')  tan  d  = 

—  92.02  tons  =  —  (M). 

Again,  by  moments: 

//7.x       5  X  773  X33  +  3X72  —  7  X  2.73 

{Ak)  = ^—^ — — ^ — ^ ^  =  +  92.02  tons. 

The  chord  stresses  resulting  from  the  upward  pressure 
alone  still  remain  to  be  found. 

The  total  negative  reaction,  supposing  the  ends  to  be 
latched  down,  has  been  shown  to  be  —  (3.53  +  3.21)  =  —  6.74 
tons.  A  margin  of  safety,  however,  of  two  tons  will  be  taken  ; 
t.  e.,  it  will  be  assumed  that  the  total  upward  pressure  at  each 
end  of  the  bridge  has  a  value  of  8.74  tons. 

In  the  example,  and  in  all  cases  where  two  or  more  systems 
of  triangulation  have  a  common  point  of  support  at  the  ends, 
some  ambiguity  necessarily  arises  in  regard  to  the  upward 
pressure.  The  proportion  of  the  excess  carried  by  either  sys- 
tem is  indeterminate  ;  and  if  there  is  no  excess,  the  propor- 
tion of  the  upward  pressure  carried  by  either  system,  during 
partial  loading  of  one  or  both  arms,  is  also  indeterminate. 


ENDS  LIFTED. 


219. 


In  the  absence  of  anything  better,  it  will  be  assumed  that 
the  excess,  in  the  example,  of  two  tons  is  equally  divided  be- 
tween the  two  systems.  It  will  farther  be  assumed  that  the 
upward  pressure,  under  all  circumstances  of  loading,  is  4.53 
tons  for  the  system  ADdEf,  etc..  and  4.21  tons  for  the  system 
ADeFg,  etc.  The  chord  stresses  due  to  the  upward  pressures 
will  then  be  the  following: 

{Ad)  —  —  8.74  X  tana  —  —     7.10  tons. 

{de)     —  {Ad\  —  4.53  X  ta?ia  =  —  10.78 

(ef)     =  (de)     —  2  X  4.21  X  tana  =  —  17.62 

{/£-)    =  (</")    -  2  X  4.53         "  =  -  24.98 

(^^^)    —  {/£')    —  4-21  {tana  +  tanS)  =  —  33.66 

{M')   =  {gh)    -  4.53  {tana  +  tan  fi)  =  -  39.32 

{DE)  =  2  X  4.21   X  tan  a  -f  4.53  tan  a  —  -\-  10.52 

{EF)  —  {DE)  +  2  X  4.53  Y.  tana  —  +  17.88 

{EG)  =  {EF)  +  2  X  4.21          "  =  +  24.72 

{GB)  =  {FG)   +  2  X  4.53          "  =  +  32.08 

As  numerical  checks: 


{hli)  -  - 


5.74  X  72 
16 


-39-33  tons. 


{GB)  -V  4.53  X  tan  fi  +  4.21  x  tan  8  —  39.32  tons 


{hh). 


The  stresses  in  the  web  members,  existing  with  a  passing 
load,  will  next  be  found,  and  those  which  may  be  termed 
counter  stresses  will  first  receive  attention. 


System  ADdE,  etc. 

Moving  loads  at  f  and  h  : 

J^i  =  7-07  +  1.09  +  4.53  =  +  12.69  to"s. 
and  {/E)  will  be  the  first  counter  stress. 


2P=o  Sit  f, 


The  shear,  s,  in  /E.  is : 

s  =  12.69  —  7-73  =  4-9^  tons. 

.'.  {/E)  :=  —  s  X  sec  a  =  —  6.40  tons. 

.-.  {dE)  =  {s  +  2.73)  sec  a  —  +  9.92  tons. 


SIVING  BRIDGES. 
Moving  loads  at  d,  f,  and  h  : 

R^  —  12.69  +  15.14  =  +  27.83  tons. 
.y(for  AD)  =  27.83  tons. 
.'.  {AD)  —  -\-  s  y.  sec  a  —  -\-  35.9  tons. 
.*.  (dD)  —  —  s  —  —  27.83  tons. 

System  ADeF,  etc. 

Moving  loads  at  e  and  g: 

R^  =  10.94  -i-  3.74  +  4.21  =  18.89  tons.     •*•  ^P—  O  at  e,  and 
{De)  is  the  first  counter  stress. 

s  (for  De)  =  18.89  "  (3  +  2.73)  =  13.16  tons. 
.-.  (De)  =  —  s  X  sec  a  =  —  16.98  tons. 
.-.  (AD)  =  {Ri  —  2)  seca=  +  20.86  tons. 

The  main  web  stresses  existing  with  the  moving  load  are 
found  as  follows : 

System  ADdE,  etc. 

Moving  load  on  Ch'  and  at  d : 

Ri     =  15-14 +  4-53 -3-53  =       16.14  tons. 

{d£)  =  -  (19.5  +  5—  16.14)  sec  a  =  —  10.78     " 
{Ef)  =  -  \dE)  +  W  seca  =  +  I4-30     " 

Moving  load  on  Ck'  and  A/: 

R^      =16.14  +  7.07  =       23.21  tons. 

(/<7)  =  -  (2  ^F  +  2w  +  W  -  Ri)  seca=  -  36.79     " 
(G/i)  =  -  i/G)  +  W  seca  =  +  40.31     " 

Moving  load  on  C/i'  and  A/i. 

Ri  =  23.21  +  1.09  =  24.30  tons. 
{/iB)=  _  (3!^+  3W+  2IV'  -R,)sec  /3  =  -  59.58  tons. 


ENDS  LIFTED.  221 

System  ADeF,  etc. 
Moving  load  on  Ch'  and  dX  e: 

Ri  =  10.94  +  4.21  —  3.21  =  1 1.94  tons. 

(^eF)  =  —  (IV  +  lu  +   f  r"  +  3  —  Ri)  sec  a  z=  —  23.59  tons. 
{Fg)  =  —  {eF)  +  ]V'  sec  a  =+  27.11     " 

Moving  load  on  Ch'  and  eg: 

7?i  =ir  11.94  +  3.74  =  15.68  tons. 

{gB  )  =  -\2{IV^-  W  +  z£/)  +  3  -  i^i ;  sec  S  =  —  66.85  tons. 

A  few  of  the  open  draw  web  stresses  are  the  following: 

(AD)  —  —  -iysec  a  =  —     3.87  tons. 

\dD)   -       o  =         0.00     " 

{dE)    =  -  Wseca  =  -    6.45     " 

(De)    =  +  (3  +  JV)  sec  a  =  +    7.39     " 

(eF)    =  -  (3  +  IV  +  jr)seca=  -  13.84     " 
(E/)    =  +  (JV+  W')seca  =  +    9.97     '' 

It  is  unnecessary  to  give  others,  as  they  are  not  needed  ; 
only  two  of  these,  it  will  be  seen,  are  used. 

All  of  the  greatest  stresses  in  the  truss  may  now  be  written 
by  the  usual  method  of  combining  the  results  for  the  different 
cases  of  loading. 

They  are  the  following: 

(AD)  =  —      3.87  tons;   +  56.75  tons. 
(dD)   =  -    27.83     " 


(dE)   =  - 

10.78 

<< 

+    9-92 

(De)   =  - 

16.98 

(< 

+    7-39 

(eF)    =  - 

23-59 

(( 

{£/)  =  - 

6.40 

u 

+  14.30 

i/Q  =  - 

36.79 

li 

i^g)  = 

+  27.11 

UB)  =  - 

66.85 

i( 

(Gh)  = 

+  40.31 

(hB)  =  - 

59.58 

a 

222 


SWING  BRIDGES. 


{Ad)  = 

+ 

2.441 

tons  ; 

-  35-521 

tor 

{de)     = 

_i_ 

6.50 

-  38.23 

a 

kef)    = 

+ 

19.88 

-  39-70 

n 

{fg)  = 

+ 

36.50 

—  20.84 

<< 

{gh)    = 

+ 

77-15 

{M)  = 

+ 

1 12.51 

(DE)  = 

— 

7.10 

+  46.2 1 

<( 

{£F)  = 

— 

17.44 

+  49-41 

<( 

(FG)  = 

— 

37-IO 

+  38.76 

<< 

(GB)  =  -    57.52     " 

The  web  members  AD,  d£,  De,  Ef,  and  the  portions  Ag 
and  DG  of  the  chords  must  be  counterbraced.  The  same 
treatment  must  of  course  be  given  to  corresponding  members 
and  portions  in  the  other  arm. 

The  particular  form  of  truss  in  the  figure  has  been  so  chosen 
as  to  illustrate  faults  of  designs,  in  general,  in  consequence  of 
possible  ambiguity  in  the  stresses. 

If  possible,  ambiguity  should  always  be  avoided.  In  the 
present  case  it  would  have  been  far  better  to  have  had  one 
system  of  triangulation,  and  supported  the  chords  by  light 
verticals,  designed  to  resist  compression,  extending  from  the 
apices. 

Precisely  the  same  methods  of  loading  and  treatment  would 
be  used  if  there  were  two  apparent  points  of  support  above  B, 
that  point  still  existing  as  the  real  point  of  support  of  the 
truss.  In  fact,  the  same  general  observations  as  those  which 
were  made  in  the  last  portions  of  Articles  38  and  40  apply  in 
this  case  also. 

The  same  methods  of  loading  and  treatment  would  also  be 
used  if  there  were  locomotive  excess,  or  if  there  were  one  or 
more  than  two  systems  of  triangulation. 

Art.  43. — Final  Observations  on  the  Preceding  Methods. 

Although  particular  forms  of  triangulation  have  been  chosen 
for  the  various  examples  in  the  different  cases  of  swing 
bridges,  yet  the  conclusions  reached  and  the  principles  estab- 


FINAL    OBSERVATIONS  ON  THE  PRECEDING  METHODS.     223 

lished  are  perfectly  general.  They  are  applicable  to  any  form 
of  triangulation,  and  to  either  the  deck  or  through  form  of 
bridge ;  they  also  apply  whether  the  two  arms  are  of  the 
same  length  or  of  unequal  length,  the  panels  being  either  uni- 
form or  irregular.  It  is  only  necessary  to  bear  in  mind  what 
may  be  called  the  "  local"  circumstances  of  any  given  case; 
these  do  not,  however,  afTect  the  general  principles.  As  a 
single  illustration — if  the  bridge  is  of  the  "  deck"  form,  those 
web  members  which  intersect  in  the  lower  chord  take  their 
greatest  stresses  together ;  if  of  the  "  through  "  form,  those 
which  intersect  in  the  upper  chord  take  their  greatest  stresses 
together. 


CHAPTER    VII. 

CONTINUOUS   TRUSSES   OTHER   THAN   SWING   BRIDGES. 

Art.  44.  —  Formvilae  for  Ordinary  Cases  —  Reactions  —  Methods  of  Pro- 
cedure. 

On  account  of  the  doubtful  utility  of  fixed  continuous 
trusses,  and  the  extreme  rarity  of  their  occurrence  in  Amer- 
ican practice,  general  directions  and  formulae  only  will  be 
given.  It  will  be  assumed  that  the  moment  of  inertia  (/)  and 
coeiificient  of  elasticity  (E)  are  constant ;  it  will  also  be  as- 
sumed that  the  points  of  support  are  all  in  the  same  level, 
as  it  has  been  shown  in  Appendix  I  to  what  cases  the  result- 
ing formulae  apply. 

\  1  2  3  4-  S  S 

Fig.   I. 

Eq.  (17)  of  that  Appendix,  after  introducing  these  condi- 
tions, gives,  in  connection  with  the  notation  of  Fig.  i,  the 
following  equations : 

2J/1  (/i  +  /o)  +  M^k  +  ^  =  O (l). 

J/i4  +  2J/2  (4  +  4)  +  .^3/3  +  ^  =  O (2). 

M^k^2M^{k^i:)-V  MJ,-V  C=o (3). 

J/3/4 +  2.1/4(4  + 4) +  ^^54  +  ^=0 (4). 

J/44 +  2^/5(4 +  4) +  ^64  +  ^  =  0 (5). 

etc.  +  etc.         +   etc.  +      =0 

The  various  values  of  M  are  the  bending  moments  existing 
at  the  supports  indicated  by  the  subscripts ;  the  moment  at  o 
is  evidently  nothing,  since  the  truss  is  there  simply  supported. 

224 


FORMULA  FOR  ORDINARY   CASES.  225 

The  quantities  A,  B,  C,  etc.,  have  the  following  values,  as 
Eq.  (17),  of  Appendix  I,  shows  : 

A  ^y^P{K'  -  z')z  ^\h  P{1^  -  z'^z    .    .    (6). 


£  =j2P{/,'-z')z  +  j2P{/,'-^)2    .    .    (7). 

*2  *S 


C  =j2Pi/i-z^)s  +  j2P{/,'-^)z    .    .    (8). 

n  =  j^ Pi/,'  - ^)s  +  jh p{/,'  -  z')z  .  .  (9). 
^=li^P(42-.^)^  +  lip(/e2-.^)^  .  .(10). 

Etc.  =  etc.  +  etc.  .    . 

The  Eqs.  (i)  to  (5)  show,  since  the  end  moments  are  zero, 
that  whatever  the  number  of  spans,  there  will  always  be  as 
many  of  those  equations  as  there  are  unknown  bending  mo- 
ments over  the  points  of  support.  Those  moments,  there- 
fore, may  always  be  found,  and,  consequently,  the  reactions 
which  depend  upon  them.  These  reactions  are  the  main 
objects  of  search.  It  will  be  necessary,  then,  to  determine 
the  bending  moments  at  the  points  of  support. 

From  Eq.  (i) : 

Mi=--y--Mi-^ — 5^.     .     .     .    (II). 
h  h 

By  inserting  this  value  of  J/,  in  Eq.  (2),  there  at  once 
results : 

44  '  44  ' 


226  CONTINUOUS    TRUSSES. 

These  values  of  M^  and  My,  inserted  in  Eq.  (3),  give: 
\-li^A{k'rmz^l^^A-2k{k^l^B+kkC 

Mi—  -ry-i  ■ 

'2  *3  M 


+ 


(  /g  /$  /4  ) 


Again,  Eq.  (4)  gives,  after  inserting  in  it  these  values  of 
M.  and  M. 


'3  '*"'J  -"^^4 


[-  2/3'^  (A  +  4)  -  2/4- (/,  +  /s)  4-  8  (/2  +  h)  (4  +  A)  (/4  +  lt)]A  -[-  hl^  + 
4  4  (4  +  A)  (/4  +  4)]  ^  +  2  /a/3  (/4  +  /s)  C  +  /a/3/4  Z> 

»~  4/3/4/5 


•  1 6  (/,  +  4)  (4  +  4)  (4  +  4)  (4  +  4)  +  44=  (4  +  4)  (4  +  4)  +  ) 
-iT/,  ^     4  4'-(4  +  4)  (4 + 4) + 44-  (4  +  4)  (4  +  4)  -  4=  4'  >  .   (14). 

Any  bending  moment  may  thus  be  found. 

It  is  seen  that  all  moments  are  given  in  terms  of  M-^,  which 
is  still  unknown.  However,  the  bending  moment  at  the  other 
free  end  of  the  truss,  from  o.  Fig.  i,  will  be  zero;  conse- 
quently its  general  expression,  put  equal  to  zero,  will  give  ^J/i 
in  terms  of  A,B,C,  etc.,  and  the  lengths  of  the  different  spans, 
/.  e.,  in  terms  of  known  quantities.  When  AI^  is  known,  all 
the  other  bending  moments  are  at  once  given  by  Eqs.  (i  i), 
(12),  (13),  (14),  etc. 

As  an  illustration,  if  there  are  five  spans,  M^  —  o  and  Eq. 
(14)  will  at  once  give  M^.  Eqs.  (11),  (12)  and  (13)  then  give 
the  other  moments  desired. 

Another  method  may  be  followed  by  which  a  less  number 
of  equations  will  suffice  for  a  greater  number  of  spans.  For 
example,  the  Eqs.  (11),  (12),  (13)  and  (14),  v/ith  a  similar 
value  for  J/g  are  aufficient  for  the  solution  of  a  case  of  ten 
spans. 

Let  A'  be  the  quantity  corresponding  to  A,  which  would 
appear  in  the  equation  involving  J/9  and  J/g,  in  a  continuous 


FORMULA.    FOR    ORDINARY   CASES.  227 

truss  of  ten  spans,  and  corresponding  to  Eq.  (i).  Let  B',  C\ 
D'  represent  similar  quantities  in  equations  corresponding^  to 
Eqs.  (2),  (3)  and  (4).  The  following  five  equations  may  then 
be  written  by  the  aid  of  Eqs.  ( i )  to  (5): 

2M,  (/lo  +  4)  +  Vl/8/9  +  A  ^o  .  .  (15) 

J/9/9+  2i^/8(/9  + 4)  +  J/7/8  +  i5'  =  o  .  .  (16) 

J/8/8+  271/,  (4  +  /;)  +  J/eA  +  6"  =  0  .  .  (17) 

M^l^^  2.\U{1.,  +  U)  +  AhU^  D-  ^o  .  .  (18) 

Ms/e  +  2  J/5  (4  +  4)  +  J/4  4'+  £    =0  .  .  (19) 

A  value  for  J/5  may  be  written  by  changing,  in  Eq.  (14), 
J/i  to  J/9,  4  to  /,o,  4  to  4,  4  to  4,  4  to  4,  4  to  4,  A  to  A',  B  to 
B\  C  to  C\  and  D  to  D'.  A  value  of  J/g  in  terms  of  J/j 
would  be  equal  to  the  value  of  M^,  given  by  Eq.  (13),  with 
exactly  the  same  changes  made,  in  so  far  as  the  same  quan- 
tities appear.  These  pairs  of  values  of  the  two  quantities  J/5 
and  J/g,  equated,  would  give  two  equations  from  which  Afi 
and  J/y  could  be  immediately  deduced.  All  the  other  mo- 
ments would  then  follow. 

The  Eqs.  (ii),  (12),  (13)  and  (14),  are  sufficient  in  them- 
selves for  the  solution  of  a  case  of  nine  spans,  in  the  manner 
just  indicated. 

The  preceding  operations  represent  the  most  direct  method 
of  finding  the  bending  moments  over  the  points  of  support. 
All  things  considered,  it  is  probably  as  short  as  anything  that 
can  be  derived. 

Prof.  Merriman  has,  however,  given  a  more  elegant  method 
by  the  use  of  so-called  "  Clapyronian  numbers."  Any  method 
involves  sufficient  tedium. 

The  preceding  formulae  will  be  very  much  simplified  if  a 
single  weight,  only,  rests  upon  some  one  span,  since  all  the 
quantities.  A,  B,  C,  etc.,  except  two,  will  then  disappear. 

The  various  reactions  may  be  immediately  determined  by 
Eqs.  (2i)-(27)  of  Appendix  I.,  after  the  bending  moments  are 
found.  And  when  the  reactions  are  known,  the  stresses  in 
the  individual  members,  for  a  given  condition  of  loading,  are 
found  precisely  as  for  a  simple  truss  supported  at  each  end. 


228  CONTINUOUS    TRUSSES. 

If  the  ends  of  the  truss  are  not  simply  supported,  the  end 
moments  must  be  known,  else  the  problem  will  be  indeter- 
minate. In  such  a  case  the  preceding  methods  are  in  nowise 
changed,  but  the  end  moments,  instead  of  being  zero,  will 
appear  as  known  quantities. 


CHAPTER   VIII. 


ARCHED    RIBS, 


Art.  45. — Equilibrium  Polygons. 

Preliminary  to  the  specific  treatment  of  arched  ribs  it 
will  be  necessary,  first  to  consider  some  general  principles  re- 
garding equilibrium  polygons  for  any  given  system  of  vertical 
forces,  and  then  those  involved  in  the  theory  of  flexure. 

In  the  figure  below,  let  AB K  h&  any  straight, 

simple  beam,  subjected  to  the  action  of  the  vertical  forces  B, 
C,  D,  etc.  Let  x  be  measured  from  any  section  positive  and 
horizontal  toward  A,  and  letP  signify  any  external  force  such 
as  the  reaction  at  A  or  any  of  the  forces  applied  to  the  beam  ; 


Fig.  I. 

then  will  2Px  represent  the  bending  moment  to  which  the 
beam  is  subjected  at  the  section  denoted  by  x.  Now  let  there 
be  imagined  any  force  /^acting  parallel  to  AEK,  and  let  the 
moments  2Px  be  taken  at  each  of  the  points  B,  C,  D,  E,  F, 
G,  H.  Then  if  the  quotients  of  those  moments  divided  by 
the  horizontal  component  of  T  be  supposed  represented  by 
the  vertical  lines  bB,  cC,  dD,  etc.,  respectively,  will  the  poly- 
gon AbcdefghK  he  one  equilibrium  polygon  for  the  given  sys- 
tem of  loads  ;  so  that  if  the  beam  AK  were  displaced  by  a  tie 
in  which  exists  the  stress  T,  and  the  given  loads  hung  from 
the  joints  b,  c,  d,  etc.,  the  whole  system  would  be  in  equilib 
rium. 

229 


230  ARCHED  A'IBS. 

In  order  to  establish  this,  it  is  only  necessary  to  show  that 
no  piece  of  the  polygon  is  subjected  to  bending;  for  if  that 
is  the  case,  the  line  of  action  of  the  resultant  stress  must  co- 
incide with  its  centre  line. 

Consider  any  portion  of  the  system,  as  that  lying  on  the  left 
of  the  vertical  line  dD.  Those  forces  which  have  moments 
about  the  point  d  are  the  external  forces  to  the  left  oidD  and 
the  stress  T  in  the  tie  AK ;  the  latter  has  a  lever-arm  «,  equal 
to  the  normal  distance  from  ^  to  AK,  and  its  moment  is  op- 
posite in  sign  to  2Px.  Consequently  the  resultant  moment 
about  d  will  h&  M  =  2Px  —  Tn.  Btit  by  construction  2Px  = 
Tn,  hence  M  =  o;  and  the  same  is,  of  course,  true  of  every 
other  joint.  If  T^  is  the  horizontal  component  of  7\  then 
evidently  Tn  =  7;  (dD)  =  2:Px. 

If  t'  be  the  general  representative  of  the  vertical  ordinates 
dB,  cC,  etc.,  then,  in  general, 

2Px 

^  h 

but  7"^  is  a  constant  quantity.  From  these  considerations 
follows  this  important  principle  : 

The  vertical  ordinates  of  the  cqiiilibritim  polygon  of  any  sys- 
tem of  vertical  loads  arc  proportional  to,  and  may  represent,  the 
bending  moments  found  at  the  various  sections  of  a  beam  sub- 
jected to  the  action  of  the  same  system  of  loads,  and  having  the 
same  span. 

Since  the  stress  T  was  taken  arbitrarily,  it  is  evident  that 
there  may  be  an  indefinite  number  of .  equilibrium  polygons 
for  any  given  system  of  loads  ;  the  principle  stated  above, 
however,  is  perfectly  general,  and  is  true  for  all. 

Since  vTj^—  ^Px  —  constant  for  any  given  section,  it  fol- 
lows that  any  variation  of  T,  and  therefore  Tf,,  produces  an 
opposite  kind  of  variation  in  7'.  Hence  the  height  of  an  equi- 
librium polygon  is  proportioned  to  the  reciprocal  of  T  or  Tj,. 

The  method  of  constructing  the  equilibrium  polygon  given 
above  is  not  the  most  convenient,  nor  the  one  commonly 
used.     The  method  ordinarily  used  is  the  usual  one  for  con- 


EQUILIBRIUM  POLYGONS. 


231 


structing  the  equivalent  polygonal  frame,  and  is  the  follow- 
ing: 

Let  AK,  Fig.  2,  represent  any  span,  inclined  in  this  case 
but  ordinarily  horizontal,  and  i,  2,  3,  4,  etc.,  the  vertical  loads 
acting  along  their  respective  lines  of  action.  In  Fig.  3  let  the 
portions  i,  2,  3,  4,  5,  6,  7,  8,  and  9  of  the  vertical  line  1-9  rep- 
resent those  loads  taken  by  any  assumed  scale.     Since  BC 


Fig.  2, 

represents  the  sum  of  all  the  applied  loads,  it  is  also  equal  to 
the  sum  of  the  two  reactions  or  shearing  stresses  at  A  and  A. 
In  the  case  of  the  simple  beam  taken,  those  quantities  will  of 
course  be  determined  by  the  law  of  the  lever  only. 

Suppose  A' C  and  A' B  to  represent  the  shearing  stresses  or 
reactions  at  ^  and  A"  respectively.  Then  draw  y^'P  parallel 
to  AK,  and  on  it  take 
any  point  P.  From  P 
draw   the    radial    lines 

a,  b,  c,  d, /,  as 

shown,  and  starting 
from  A  or  K  in  Fig.  2, 
draw  the   lines  a,  b,  c, 

d, /,    parallel 

to  the  lines  denoted 
by  the  same  letters  in 
Fig.  3.  Then  will  Fig. 
2  represent  the  equi- 
librium polygon  for  the 
given  span  and  system 
of  loading. 

The  line  AK  ox  PA' 
is  called  \}!\q.  closing  line  of  the  polygon.     The  reaction  at  A  is 


232  ARCHED  RIBS. 

evidently  composed  of  the  numerical  sum  of  the  vertical  com- 
ponents in  /  and  AK,  while  that  at  K  is  equal  to  the  numer- 
ical difference  of  the  vertical  components  in  a  and  AK. 

The  point  P,  from  which  the  radial  lines  are  drawn,  is  called 
the  pole,  and  the  normal  distance  from  the  pole  to  the  load 
line  BC,  t\\Q  pole  distance.  The  pole  distance  evidently  rep- 
resents the  horizontal  component  of  stress  common  to  all  the 
members  of  the  polygon. 

In  order  that  the  equilibrium  polygon,  constructed  accord- 
ing to  the  principles  given  above,  shall  exactly  fit  the  span,  it 
is  only  necessary  that  a  proper  observance  be  paid  to  the 
scales  used. 

2Px 
From  the  equation  7^  =  — = — it  is  seen  that  the  scale  for  the 

■I  h 

forces  does  not  affect  the  height  of  any  joint  of  the  polygon  ; 
it  depends  only  on  the  scale  according  to  which  x  or  the  hor- 
izontal span  is  drawn. 

Let  the  line  PP'  be  drawn  parallel  to  EC,  and  let  P'  be 
the  pole  of  a  new  equilibrium  polygon  ;  the  pole  distance 
will,  of  course,  remain  the  same  as  before.  But  the  pole  dis- 
tance represents  the  horizontal  component  of  the  stress  in  the 
closing  line,  and  it  has  already  been  shown  that  if  7)^  remains. 


the  same,  7>  cannot  vary.     Hence,  any  movement  of  the  pole 
parallel  to  the  load  line  does  not  chafige  the  vertical  dimensions 


EQUILIBRIUM  POLYGONS. 


23: 


■of  the  equilibrium  polygon.     But  if  the  pole  distance  is  changed, 
the  vertical  di>nensions  are  changed  in  the  inverse  ratio. 

The  determination  of  the  deflection  polygon  of  an  arched 
rib  with  ends  fixed,  involves  the  use  of  an  equilibrium  poly- 
gon, similar  to  that  required  for  a 
system  of  forces  whose  resultant  is 
a  couple.  Its  method  of  construc- 
tion is  not  at  all  different  from 
that  just  given. 

In  Fig.  4,  let  the  forces  1,2,  3, 
4.  5.6,  7,  8,  and  9,  act  vertically, 
BC  being  horizontal,  and  let  the 
sum  of  I,  2,  3,  8,  and  9  be  numeri- 
cally equal  to  the  sum  of  4,  5,  6, 
and  7.  The  double  line,  DE,  in 
Fig.  5,  represents  the  forces  shown 
in  Fig.  4. 

In  Fig.  5,  draw  the  line  rt:  in  a 
horizontal  direction    through  the 
upper  extremity  of  force   i,   and 
take  any  point  on  it  for  the  pole  P. 
lines  in  the  usual  manner  as  shown. 

From  C,  in  Fig.  4,  draw  b'  parallel  to  b  in  Fig.  5,  until  it  in- 
tersects the  line  of  action  of  force  2.  Then  draw  the  other 
lines,  ^',  d\  etc.,  parallel  to  r,  d,  etc.,  until  the  lines  of  action 
of  the  other  forces  are  intersected,  b' ,  c\  .  .  .  .  h' ,  k',  will  then 
be  the  equilibrium  polygon  for  the  system  of  forces  assumed. 

It  is  seen  that  the  polygon  does  not  close.  This  simply 
shows  that  the  resultant  of  the  system  is  a  couple,  whose  mo- 
ment is  the  force  a,  in  Fig.  5,  multiplied  by  AB  (vertical)  in 
Fig.  4. 

The  following  general  principle  then  results  :  The  equilib- 
rium polygon  for  any  system  of  parallel  forces  zvhose  resultant 
is  a  couple,  is  not  a  closed  o?te. 

This  principle  is  indeed  true  for  any  system  of  forces. 

P  might  have  been  taken  at  any  other  point,  as  P'  in 
OP.  In  that  case,  however,  the  equal  forces  a,  acting  at  A 
and  C,  Fig.  4,  would  be  parallel  to   a  line  drawn  from  the 


Fig.  5. 
From  P  draw  the  radial 


n  t  4 

-o4 


ARCHED  RIBS. 


upper  extremity  of  force  i  to  P' .  The  vertical  dimensions  of 
the  polygon,  measured  from  either  of  the  forces  a  (in  gen- 
eral inclined),  will  always  be  the  same  if  the  pole  remains  in 
the  linePP'. 

Art.  46. — Bending  IVIoments. 

An  arched  rib  is  any  truss  curved  in  a  vertical  plane,  both 
of  whose  chords  are  convex  or  concave  in  the  same  direction, 
neither  being  horizontal ;  the  ends  may  be  fixed  or  free. 

In  Fig.  2  of  Art.  45,  let  ADEK  represent  an  arched  rib 
sustaining  the  loads  i,  2,  3,  4, ..  .  .  .  9.  Now  it  has  already 
been  seen  that,  so  far  as  equilibrium  is  concerned,  any  given 
system  of  loading  may  be  sustained  by  any  one  of  a  set  of 
equilibrium  polygons  consisting  of  an  indefinite  number.  On 
the  other  hand,  it  is  evident  that  no  polygon  or  arched  rib 
can  be  drawn,  which  is  not  an  equilibrium  polygon  for  some 
system  of  vertical  loading;  but  if  that  arched  rib  sustains 
some  other  system  of  loads  than  that  which,  it  may  be  said, 
properly  belongs  to  it,  and  if  its  joints  be  prevented  from 
turning,  it  will  be  subjected  to  bending,  which  will  vary  from 
one  section  to  another. 

The  arched  rib  ADEK  sustains  a  system  of  loading  for 
which  AfK  is  the  equilibrium  polygon,  hence  the  former  will 
be  subjected  to  varying  degrees  of  bending  at  various  sec- 
tions. When  the  rib  ADK  is  subjected  to  the  action  of  its 
load,  stresses  are  developed  in  its  different  parts,  whose  hori- 
zontal components  are  all  the  same  because  the  load  is  wholly 
vertical.  Now  if  an  equilibrium  polygon  can  be  found  in 
which  the  horizontal  component  of  stress  Tj,  is  the  same  as 
that  developed  in  ADEK,  then  all  the  circumstances  of  stress 
and  bending  in  the  latter  can  be  determined,  as  will  be  seen 
hereafter. 

Suppose  AfK  to  be  that  polygon,  then  let  v'  denote  the 
portion  of  a  vertical  line  intercepted  between  it  and  the  arched 
rib,  as  DD' .     The  moment  about  any  point  D  will  then  be 

M  ^^Px-  T„  (7'  4^  v). 

But  since  AfK  is  the  equilibrium  polygon,  '^  Px  —  Tf^v  —  O. 


RENPIXG  MOMEXTS. 


235 


.'.  M  =  —  T^  v'.  When  the  polygon  lies  above  the  rib,  v'  is 
negative,  and,  hence,  M  positive. 

Let  the  polygon  which  has  the  same  value  of  7\  as  the 
arched  rib  be  called  the  true  equilibrium  polygon  ;  then,  since 
Th,  is  a  constant  quantity  for  the  same  rib,  there  is  established 
the  following  important  principle: 

J  lie  bending  moj/ients  to  zvJiich  the  different  parts  of  an  arched 
rib  are  subjected  are  proportional  to,  and  may  be  represented  by, 
the  vertical  intercepts  included  between  the  rib  and  the  triie 
eqiiilibriuni  polygon. 

This  principle  has  been  demonstrated  for  a  beam  with  free 
ends  only,  but  it  is  true  also  for  a  beam  with  fixed  ends,  as 
will  now  be  shown. 

In  order  to  fix  the  end  of  the  rib  it  is  only  necessary  to  im- 
press upon  the  rib  at  A,  the  point  of  fixedness,  the  proper 
couple  whose  moment  is  in ;  in  the  fixed  rib,  as  in  the  free, 
let  Tj^  represent  the  horizontal  thrust.  The  true  equilibrium 
polygon  for  the  fixed  rib  will  be  that  found  by  increasing  the 
vertical  dimensions  of  a  polygon  for  a  free-end  rib,  formed  by 
vising  Tf^  and  reactions  for  fixed  ends,  by  a  constant  amount 

m 
equal  to  -7fr  .      Again,  taking  moments  about  any  point  of 

the  centre  line  of  the  rib,  there  will  result 

M—^ Px  +  ;//  —  7;  (f  +  ™-  +  t^)  =  —  Tj,v', 

^  h 

as  before.  This  shows  that  the  principle  stated  above  is  true 
fpr  both  fixed  and  free-end  ribs.  In  truth  this  equation  might 
have  been  written  first,  and  the  special  case  of  the  free-end 
rib  deduced  by  making  m  =  o. 

An  arched  rib,  then,  when  subjected  to  the  action  of  a  load, 
suffers  bending  in  the  same  manner  as  a  straight  beam,  but  to 
a  different  degree. 

In  so  far  as  it  plays  the  part  of  a  beam,  it  must  be  governed 
by  the  general  laws  of  bending  or  flexure.  The  formulae  to 
be  given  in  this  connection  are  those  approximate  ones  based 
on  the  common  theory  of  flexure,  and  found  in  the  ordinary 
works  on  that  subject. 


:236  ARCHED  RIBS. 


Art.  47. — General  Formulae. 

Let  ^^  denote  the  total  shearing  stress  at  any  section,  P  an/ 
applied  load  or  external  force,  M  the  bending  moment  at  any 
section,  and  D  the  deflection  found  above ;  then  the  six  gen- 
eral equations  of  flexure  demonstrated  in  the  Appendix  on 
the  Theorem  of  Three  Moments,  some  of  which  are  made  use 
of  in  the  graphical  treatment  of  arched  ribs,  are  the  following: 

5  =  ^P, 

M        =:2Px, 

p,        _M^ 
EI  ' 

nM 

D  =2  nP  X  =  2.      r-T  ■  • 

EI 

D„        =^nPy. 

If  the  beam  is  originally  straight  and  parallel  to  the  axis  of 
X,  n  becomes  dx  and  y  =  o. 

As  usual,  E  and  /  represent  the  coefficient  of  elasticity  and 
moment  of  inertia  of  the  cross-section,  respectively. 

The  limits  of  the  summations  are  the  section  considered 
and  any  section  of  reference.  The  quantities  S,  M,  P'  and 
D  then  refer  simply  to  that  portion  of  the  beam  over  which 
the  summation  extends. 

One  very  important  deduction  is  to  be  drawn  from  the  above 
equations,  or  rather  from  the  second  and  fifth  of  them.     It  is 

seen  from  these  two  equations  that  -^^^  stands  in  the  same 
relation  to  D  that  P  does  to  M.     Consequently  if  -ry  be  put 


ARCHED   RIB    IVITH  ENDS  FIXED. 


237 


/n  the  place  of  P  in  any  graphical  construction,  D  will  be 
represented  in  the  place  of  M.     If,  therefore,  an  "  equilibrium 

polygon  "  be  constructed  for  any  span  by  taking  -^^  as  loads 

instead  of  P,  the  vertical  ordinates  of  the  polygon  will  repre- 
sent the  deflections  at  the  sections  denoted  by  the  corresponding 
values  X. 

This  polygon  may  be  called  the  "  deflection  polygon,"  and 
its  construction  plays  a  very  important  part  in  the  determina- 
tion of  the  true  equilibrium  polygon  for  an  arched  rib,  in  the 
majority  of  cases. 


Art.  48. — Arched  Rib  with  Ends  Fixed. 

The  ends  of  an  arched  rib  or  any  girder  are  considered 
fixed  when  the  angles  formed  by  their  centre  lines  with  any 
assumed  line,  at  the  fixed  sections,  do  not  vary  under  any 
applied  load. 

Let  BAD,  PI.  v.,  be  the  centre  line  of  any  arched  rib  ;  it 
will,  of  course,  be  considered  fixed  at  the  points  B  and  D. 
This  line  may  be  any  curve,  though  for  convenience  the  arc 
of  a  circle  has  been  drawn.  In  the  demonstration  no  atten- 
tion whatever  has  been  given  to  the  character  of  the  curve, 
so  that  it  is  equally  applicable  to  any  other  curve. 

In  the  present  case  the  centre  line  of  the  rib  will  be  divided 
into  equal  parts  for  the  application  of  the  load,  and  each  of 
those  parts  will  be  n  ;  consequently  that  quantity  will  have  a 
finite  value,  and  the  results  will  not  be  strictly  accurate, 
though  near  enough  for  all  technical  purposes. 

The  piers  or  points  of  fixedness  are  supposed  to  be  immov- 
able, whatever  may  be  the  character  of  the  load  ;  but  if  that 
is  the  case,  the  summation  of  the  strains  at  any  given  dis- 
tance from  the  neutral  axis  of  the  rib,  considered  as  a  truss, 
taken  throughout  the  whole  length  of  the  rib  BAD,  must  be 
equal  to  zero.  Take  that  distance  as  unity,  then  there  re- 
sults, for  one  condition,  since  71  is  constant,  the  equation  : 


258  ARCHED   RIBS. 

^  nP'  =  :^  ^^  =  —-:e  m  =  o. 
"EI       EI    " 


^..M=o. 


The  quantity -=-=  is  brought  outside  of  the  sign  ^\  because 
El 

the  moment  of  inertia  of  the  cross-section  of  the  rib  is  s7ipposed 

to  be  the  same  throughout  its  entire  length.     More  will  be  said 

on  this  point  hereafter. 

It  has  already  been  shown  that  the  area  included  between 
the  equilibrium  polygon  and  the  curve  BAD  is  made  up  of 
vertical  strips,  whose  lengths  (the  vertical  intercepts)  repre- 
sent the  actual  bending  moments  at  the  different  sections  of 
the  rib.  Hence  2M  represents  the  sum  of  those  vertical 
lengths  or  intercepts  drawn  at  the  points  to  ivhich  the  inonients 
M  belong,  and  the  equation  2  M  =^  o  shows  that  the  sum  on 
one  side  of  BA  D  must  be  equal  to  that  07i  the  other. 

But,  as  will  be  seen,  there  may  be  an  indefinite  number  of 
equilibrium  polygons  which  will  fulfill  this  condition  ;  conse- 
quently at  least  one  other  condition  must  be  obtained. 

Since  the  points  B  and  D  are  fixed,  the  sum  of  all  the  de- 
flections, both  horizontal  and  vertical,  taken  between  those 
two  points,  must  be  equal  to  zero.  It  has  been  shown  that 
the  vertical  deflection  at  any  point,  when  n  and  /  are  con- 

n 
sidered  constant,  is  Z>  =  —  -  '2Mx ;    also,  from  the  reasoning 
•  El 

applied  to  the  curved  girder,  that  the  horizontal  deflection  is 

n 
Dh  —  -jrj  2My.     Now,  when  these  summations  extend  from 

^to  D,  since  those  points  are  fixed,  both  D  and  D,^  must  equal 
zero.  The  three  equations  of  condition,  then,  which  must  be 
fulfilled  for  the  rib,  are : 

2lMy=  o. 
It  has  already  been  stated,  and  it  is  evident  without   much 


ARCHED  RIB    WITH  ENDS  FIXED.  230 

thought,  that  any  polygon  whatever  is  an  equihbrium  polygon 
for  some  load. 

Hence  consider  i^^'^Z?,  PL  V.,  an  equilibrium  polygon  for  its 
proper  load,  and  consider  it  subjected  to  that  load  ;  denote  its 
moments  by  M^,. 

Again,  suppose  the  polygon  a,  a ,  a",  etc.,  to  be  the  true 
equilibrium  polygon  for  the  given  load,  and  denote  its  mo- 
ments, represented  by  the  vertical  ordinates  drawn  from  its 
closing  line,  by  M^. 

Then,  from  the  principle  which  precedes  the  equation  M  = 
—  Tf{v'  in  the  general  discussion  of  equilibrium  polygons, 
there  follow  the  equations  : 

.'.  ^]Mx  =  ^]M,x  -  ^]M,x  =  o. 
Or,  ^"Jlax  =  ^"Jhx. 

In  the  same  manner,  ^'^^Maj^  =  '^'^^M^.y.  This  last  equation 
will  be  used  in  fixing  the  pole  distance  of  the  true  equilibrium 
polygon. 

It  must  be  remembered  that  M  represents  the  actual  mo- 
ment to  which  the  rib  is  subjected  at  any  point. 

The  application  of  these  two  conditions  will  be  shown  in 
the  course  of  the  construction  of  the  true  equilibrium  poly- 
gon, as  they  are  needed. 

In  the  figure  of  PI.  V.,  let  the  scale  for  linear  measurements 
be  10  feet  to  the  inch,  and  the  force  scale  15  tons  per  inch. 
The  curved  centre  line  of  the  rib  is  divided  into  ten  equal 
parts  of  10.95  feet  each,  and  that  is  the  constant  value  of  n. 
The  load  is  not  therefore  uniformly  distributed.  The  panel 
length,  or  horizontal  distance  between  the  points  of  appli- 
cation of  the  loads,  is  thus  a  variable  quantity.  If  the  versed 
sine  of  the  centre  line  of  the  rib  is  small,  n  may  be  taken 
equal  to  the  span  divided  by  the  number  of  panels.  But  if 
the  versed  sine  is  not  larger,  even,  than  in  the  present  case,  n 


240 


ARCHED  RIBS. 


cannot  be  so  taken  without  sensible  error,  as  will  be  seen. 
The  other  data  are  as  follows : 

Span  =  loo  feet. 

Radius  =    75  feet. 

Angular  length  of  curve  =  83^  'i^'j. 
Panel  fixed  load  =    4  tons. 

Panel  moving  load  =  10  tons. 

Centre  rise  of  rib  =  19.  i  feet. 

In  the  figure  BD  is  the  span,  and  C  the  centre  ;  b\  b",  b"\ 
etc.,  are  the  panel  points  equidistant  in  the  curve,  and  through 
which  the  loads  are  supposed  to  be  applied. 

Now  the  actual  moment  area  for  the  arched  rib  is  supposed, 
really,  to  be  the  difference  between  the  moment  area  of  the 
true  equilibrium  polygon  for  the  applied  loads  and  that  of  the 
rib  itself  considered  as  an  equilibrium  polygon  for  the  proper 
load  ;  both  systems  of  loading  being  supposed  applied  to  a 
straight  beam  fixed  at  each  end. 

The  first  portion  of  the  problem  which  presents  itself,  then, 
is  to  determine  the  true  equilibrium  polygon  for  the  given 
load.  The  construction  will  first  be  made,  and  it  will  then  be 
shown  that  the  two  conditions  given  above  are  satisfied. 

Let  the  moving  load  cover  the  left  half,  BC,  of  the  span, 
and  suppose  the  half  panel  loads  at  B  and  D  to  rest  directly 
on  the  abutments.  According  to  the  scale  taken,  lay  off  B6 
equal  to  7  tons,  half  the  total  load  on  the  panel  Ab'^"' ,  and 
B^  equal  to  half  the  fixed  panel  load  on  Ab"^;  then  lay  off 
6—  10,  divided  into  four  equal  parts,  equal  to  the  four  equal 
panel  loads  on  b\  b",  b'" ,  and  b^''.  In  the  same  manner  lay 
off  5  —  I,  equal  to  the  four  fixed  panel  loads  on  the  right  half 
of  the  span.     Assume  6"  as  a  convenient  pole,  and  draw  the 

radial  lines  from  it  to  i,  2,  3,4, 10.     Starting  from  C, 

draw  C  —  6 until  it  intersects  a  vertical  through  ^'^  at  Ui;  from 
the  latter  point,  a^  a^  parallel  to  Cy  until  it  intersects  the  ver- 
tical through  b'"  ;  proceed  in  the  same  manner  until  the  poly- 
gon ECF  is  drawn.  If  the  ends  were  free  EF^owld  be  the 
closing  line,  but  one  must  now  be  found  that  will  satisfy  the 


ARCHED  RIB    WITH  ENDS  FIXED.  24 1 

condition  ^  Ma  =  o,  or,  in  other  words,  the  sum  of  the  verti- 
cal intercepts  drawn  from  the  closing  line  downwards  must 
be  just  equal  to  the  sum  of  those  drawn  from  the  same  line 
upwards. 

The   proper    closing   line    is   easily    located    by   trial.       If 
6V=  0.5  inch  and  vv'  =  0.96  inch,  there  results: 


.B 


Til 


2^  Mf,  =  ^vv'  +  a-^c  —  U'f"  —  a^c'"  —  a^c^''  —  Cc^  —  a^f^  —  a-f 
+  ^8*^^"  +  ^9^*^  +  i^'V"  =  1.88  -  1.89  =  -  o.oi  inch. 

This  sum  is  sufficiently  near  zero. 

The  lines  vv'  and  v"v"'  are  drawn  vertically  through  points 

n 
b  and  h^,  distant  —  from  B  and  D  on  the  curve  BAD,  and 

4 
their  halves  are  taken  because  in  the  summation  2   iiMa  there 

1  ,  ,  n  ,11         ,,    ,,,  vv' 

would   appear  terms  —  x  vv    and  ~  x  v  v   ,  or  n  x  —  and 

v"v"' 
n  X 


2 

Similar  terms  will  hereafter  appear  in  similar  summa- 
tions. 

The  closing  line  HK  then  satisfies  the  condition  2 ^AI^  —  o 
for  the  equilibrium  polygon  ECF. 

There  still  remains  the  condition  2  J/^x  =  2  M,,x.  This 
equation  will  be  satisfied  by  making  each  of  its  members 
equal  to  zero.  The  closing  line  HK  must,  then,  also  make 
2^MaX  =  o.  This  simply  means  that  the  vertical  ordinates 
of  the  polygon  a,  measured  from  HK,  multiplied  by  their 
horizontal  distance,  from  D,  will  form  a  sum  equal  to  zero 
when  their  products  are  added.  If  the  ordinates  below  HK 
are  taken  positive,  as  v"v"' ,  agC^^.  a^c",  etc.,  and  those  above, 
as  a^c"^,  negative,  and  if  the  ordinates  and  distances  be  taken 
by  scale  from  the  drawing,  there  will  result,  nearly, 

v"v"'  X  De  +  a^c^  x  De'  +  a-^c'  x  De^  +  etc.  =  +    99.7 

Orfi^  X  De"  +  a^c^  xDe"+  etc.  =  —  102.0 

16 


242  ARCHED    RIBS. 

The  numerical  values  are  nearly  enough  equal,  and  the  line 
HK  will  be  taken  as  the  proper  closing  line. 

The  next  step  is  to  find  the  closing  line  for  the  curve  BAD 
of  the  rib,  considered  as  an  equilibrium  polygon,  which  will 
satisfy  the  same  general  conditions. 

Using  precisely  the  same  method  of  procedure  as  for  the 
pohgon  ECF,  the  line  H" K  '  is  found  to  be  the  one  desired, 
for  that  line  makes  the  sum  of  the  intercepts  above  it  just 
equal  to  the  sum  of  those  below  it. 

Ac^  is  about  0.62  of  an  inch. 

For  this  curve  the  summation  may  be  written  : 

2^J/j  =  bH'  -f-  2<5'Vi  +•  2b"C'i,  —  2c^b"'  —  ic^b^"  —  Ac^  =  o. 

Since  the  curve  is  symmetrical  in  reference  to  A,  the  static 
moments  of  the  ordinates  on  one  side  of  H" K'\  about  DK" , 
will  evidently  be  equal  to  the  same  moment  of  those  on  the 
other  side. 

The  second  condition  may  now  be  applied.  That  condition 
is  2^May  =  -S^il/ftj.  It  has  already  been  shown  that  if  M^ 
and  Jlfi,  be  considered  loads  applied  at  distances  y  from  the 
assumed  origin,  the  ordinates  of  the  equilibrium  polygon  so 
constructed   will    represent    the    quantities    2  Maj  = -D^,  or 

Through  A,  therefore,  draw  the  horizontal  line  RAS.  As- 
sume any  line,  as  AC,  as  the  closing  line  of  the  deflection 
polygon,  and  lay  ofi  At  equal  to  a  half  of  vv'.  Also  make  tt 
equal  c'a-^;  ft"  equal  d'a^^;  t"t"'  equal  a^c",  etc.  /"/'"  is 
measured  in  a  direction  opposite  to  that  of  the  preceding, 
because  it  represents  a  moment  of  an  opposite  sign.  In  the 
same  way  At  measured  to  the  right  of  A  is  equal  to  a  half  of 
v"v"' ;  tty  equal  to  ^^^9 ;  t^t,,  equal  to  «8<^^"\  etc.*  Draw 
bb'^  and  note  its  intersection  C"  with  AC.  From  C"  draw 
C"d^  parallel  to  Ct  until  it  intersects  a  horizontal  drawn 
through  b'  in  d' \  draw  d'^d''  parallel  to  Ct'  until  it  intersects 
a  horizontal  through  b"   in  d" ;  draw  d''d'"  parallel  to  Ct" 

*  If  the  drawing  had  been  made  accurately,  /'"/iy  would  have  been  exactly 

equal  to  Cc. 


ARCHED    RFB    WITH  ENDS  FIXED. 


243 


until  it  intersects  a  horizontal  through  //",  etc.  The  polygon 
C'd'd'',  etc.,  will  intersect  the  horizontal  RAS  at  a  point 
distant  from  A,  on  the  left  of  it,  from  which  /  is  drawn  to  a 
point  on  the  right  of  A  at  the  intersection  of  the  deflection 
polygon  formed  by  using  /,  /  ,  /, ,  /  ,  as  before,  and  which  is 
shown  in  the  figure.  The  sides  of  the  deflection  polygon  on 
the  right  of  CA  are  parallel  to  radial  lines  drawn  from  C  to 
the  points  /,  /,  /  ,  /,  .  The  distance  (/  —  /  )  represents,  in  an 
exaggerated  manner,  the  horizontal  deflection  of  the  end  of  a 
vertical  beam  fixed  at  C,  whose  length  is  CA,  and  which  is 
subjected  to  the  bending  moments  vv ,  ca,  c"a^,  etc.,  at  verti- 
cal distances  from  C  equal  to  the  heights  of  b,  b\  b",  etc.,  above 
BD.  In  the  same  manner,  /  represents  the  same  quantity 
for  the  same  beam  when  subjected  to  the  corresponding  mo- 
ments on  the  right-hand  side  of  A  C. 

The  line  /  can  be  determined  with  less  work  and  more  sim- 
ply when  the  meaning  of  the  construction  is  once  clearly 
seen,  by  laying  ofT,  on  the  left  of  y^,  as  before,  loads  repre- 
sented by  the  algebraic  sums  (At  +  At),  (//'  +  //,),  {t't" -\-t t }, 
etc.,  and  then  drawing  the  equilibrium  polygon  as  usual.  The 
distance  from  A  to  the  intersection  of  the  polygon  with  AR 
will  then  be  equal  to  /. 

The  deflection  polygon  d'd"d"'d^^  is  constructed  in  pre- 
cisely the  same  manner  as  the  preceding.  Make  As  equal  to 
a  half  of  bH' ;  ss  equal  to  cb'  ;  s s"  equal  to  c^^b"  ;  s's"  equal 
to  c^b"\  etc.;*  then  draw  radial  lines  from  those  points  of 
division  to  C.  The  point  d'  is  at  the  intersection  of  C"d' , 
drawn  parallel  to  Cs,  with  a  horizontal  line  drawn  through  b' ; 
d'd"  is  drawn  parallel  to  Cs  until  it  intersects  a  horizontal 
line  drawn  through  b" ,  and  the  other  sides  of  the  polygon  are 
constructed  in  the  same  way. 

The  polygon  cuts  the  horizontal  line  RAS  in  a  point  distant 
i/ from  A.  There  will,  of  course,  be  another  deflection  poly- 
gon, precisely  the  same  as  the  last,  on  the  right-hand  side  of 
AC,  found  by  taking  ^K'D,  c^b^^,  etc.,  and  laying  them  off 
from  A  towards  S. 

*  With  a  sufficient  accuracy  of  construction,  As^^  would  equal  a  half  of  Ac^ 


244 


ARCHED  RIBS. 


If  the  intercepts  used  in  the  deflection  polygons  for  Ma  and 
>/ft,  represent  those  moments  by  the  proper  scale,  then  by  the 
same  scale  CA  will  not,  in  general,  represent  the  true  pole 
distance.  But  this  fact  has  the  same  proportional  effect  on  both 
/and  /.  Consequently  any  result  depending  on  the  equality 
of  /  and  /,  will  not  be  affected. 

Instead  of  using  Acr^  and  Cc^  in  the  manner  shown,  greater 

accuracy  might  have  been  attained  by  taking  half  intercepts 

n 
at  the  distance  —  on  both  sides  of  A  and  C.     Such  an  oper- 

4 
ation,  however,  is  unnecessary  in  all  ordinary  cases,  since 
moments  in  the  vicinity  of  A  and  C  have  very  little  effect  on 
the  horizontal  dimensions  of  the  deflection  polygon.  Mo- 
ments, on  the  contrary,  in  the  vicinity  of  HH" ,  have  great 
effect. 

Now  /,  represents  2  ^  M^y  ;  and  /,  -2  ^  Mj,y ;  and  in  order 
that  the  second  condition  may  be  satisfied  they  should  be 
equal.  Since  /^  is  less  than  /,  it  shows  that  the  quantities  M^ 
are  too  small,  or,  in  other  words,  the  pole  distance  BC  is  too 
large.  This  last  statement  is  evidently  true,  if  it  be  remem- 
bered that  the  pole  distance  is  inversely  proportional  to  the 
vertical  ordinates  which  represent  the  moments. 

Lay  off,  therefore,  on  yi  (7  produced,  the  distance  6" J/ equal 
to  /i,  and  draw  through  M  the  horizontal  line  MN.  With  a 
radius  CN  equal  to  /,  draw  the  arc  of  a  circle  cutting  MN  in 
N,  then  produce  the  line  CP.  All  moments  represented  in 
the  lower  equilibrium  polygon  a  will  have  to  be  increased  in 
the  ratio  of  67V  to  CM.  To  make  this  reduction,  draw  a 
horizontal  line,  for  instance,  through  c^  until  it  cuts  CN  mp, 
then  make  c^a^  equal  to  Cp\  a^  will  be  one  point  in  the  true 
equilibrium  polygon.  All  other  points  might  be  found  in  the 
same  way,  but  having  found  one  point,  as  a^ ,  a  much  shorter 
method  may  be  used. 

It  has  already  been  shown  that  the  vertical  dimensions  of 
two  equilibrium  polygons  for  the  same  loading  and  span  are 
inversely  proportional  to  the  pole  distances,  the  vertical 
dimensions  being  measured  from  the  closing  lines.     In  the 


ARCHED  RIB    WITH  ENDS  FIXED. 


245 


figure  the  vertical  dimensions  of  the  polygon  ECF  must  be 
increased  in  the  ratio  of  Z,  to  /,  or  in  the  ratio  of  Cc^  to  Cp. 
Hence,  on  CN  produced  make  CP  equal  to  BC,  and  draw  the 
horizontal  line  OP  cutting  CM  produced  in  O,  then  will  CO 
be  the  pole  distance  for  the  true  equilibrium  polygon. 

In  order  to  find  the  true  pole,  make  CL  parallel  to  HK, 
then  draw  LC  parallel  to  BC^  and  equal  to  C0\  the  point 
C  will  be  the  true  pole. 

According  to  previous  principles,  the  reactions  or  vertical 
shearing  stresses  at  B  and  D  will  be  Z  —  10  and  L  —  i  respect- 
ively, and,  since  the  closing  line  must  be  parallel  to  BD  in  the 
true  equilibrium  polygon,  LC  must  be  parallel  to  BC.    From 

C  draw  the  radial  lines  shown  to  i,  2,  3, and  10; 

these  lines  will  be  parallel  to  the  sides  of  the  true  polygon  ; 
/.  e.,  draw  rtV^  parallel  to  C  6  until  it  cuts  a  vertical  through 
b'^'^  \  a^^a"  parallel  to  6"  7  until  it  cuts  a  vertical  line  drawn 
through  d'" ;  a^a^  parallel  to  C  5,  etc.     The  polygon  aa'a" 

a^^a^   so    formed    will    be    the    true    equilibrium 

polygon. 

As  a  check  some  of  the  points  as  a'  or  a"  should  also  be 
found  by  the  previous  method.  Thus  make  6^^  equal  to  c'a, 
and  draw  gp'  parallel  to  CD  ;  Cp'  should  then  be  equal  to  ca'. 
Other  points  may  be  treated  in  the  same  manner. 

It  may  now  be  seen  that  the  polygon  a  a  a" a^ 

satisfies  the  three  conditions  2^M=o,  2  Mx  =  o,  and 
'S^My  —  o.  The  first  conditions  are  evidently  satisfied  by 
the  method  of  locating  the  closing  lines  in  the  lower  polygon 
UiO^ rt'g,  and  the  'curve  BAD,  for  any  intercept  be- 
tween the  curve  and  the  upper  polygon  a  is  the  difference 
between  two  intercepts  each  of  which  belongs  to  a  sum  equal 
to  zero,  hence  the  sum  of  those  intercepts  is  zero.  The  second 
condition  is  satisfied  by  the  location  of  the  point  C. 

Another  check  on  the  degree  of  accuracy  attained  in  the 
construction  is  found,  in  what  has  just  been  said,  i.  e.,  the 


*  The  reason  for  the  true  closing  line  being  horizontal  is  given,  though  for 
soother  purpose,  on  page  273. 


24^3 


ARCHED   RIBS. 


moment  area  lying  above  the  curve  BAD  must  be  just  equal 
to  that  lying  below  it. 

By.  actual  measurement  LC  is  equal  to  3.65  inches,  hence 
the  constant  horizontal  component  of  stress  in  any  portion  of 
the  rib  is  54.8  tons.  The  resultant  stress  at  any  point  of  the 
arched  rib  is  equal  to  54.8  tons  multiplied  by  the  secant  of 
the  inclinations  at  that  point.  The  bending  moment  at  any 
point  to  which  the  rib  is  subjected  is  found  by  multiplying 
the  vertical  intercept  between  the  equilibrium  polygon  and 
the  curve  by  (54.8  tons  =  Tj).  For  example,  the  actual  mo- 
ment to  which  the  rib  is  subjected  at  h'  is  {b' ci   x  54.8). 

The  line  of  action  of  Tf,  (54.8  tons  in  this  case)  is,  of  course, 
along  the  true  closing  line  H"K".  This  is  an  important 
matter,  as  will  hereafter  be  seen. 

A  line  drawn  through  C  parallel  to  RF  would  cut  off  on 
the  load  line  the  reactions  which  would  exist  were  the  ends 
free. 

The  reaction  at  B,  in  the  present  case,  is  thus  seen  to  be 
much  greater  than  would  be  found  in  the  case  of  free  ends. 

A  point  in  the  vertical  line  passing  through  the  centre  of 
gravity  of  the  load  is  found  at  the  intersection  of  the  sides  a 
a  and  a^  a^  in  G.  G' ,  at  the  intersection  of  Ea^,  and  Fa^, 
prolonged,  is  in  the  same  vertical  line. 

The  pole  of  the  deflection  polygons  might  have  been  at  A, 
and  the  moments  laid  off  from  C,  in  which  case  the  half  of 
Acr,,  and  the  same  of  Cc",  would  have  been  the  moment  dis- 
tances adjacent  to  C.  Precisely  the  same  value  iov  LC  would 
probably  not  be  found,  because  the  sum  ^  My  is  not  con- 
tinuous, and  consequently  not  exact.  For  this  reason  it  would 
be  better  in  an  actual  case  to  divide  the  real  panel  lengths 
into  two  or  more  equal  parts  in  order  to  find  the  true  pole 
distance,  LC\  and  consequently  the  true  equilibrium  polygon. 
Th  can  then  be  used  to  find  the  stresses  in  the  members  of 
the  actual  rib  in  a  manner  that  will  hereafter  be  shown. 

The  diagram  should  of  course  be  drawn  to  as  large  a  scale 
as  possible,  and  it  may  often  be  advisable  to  exaggerate  the 
vertical  scale  so  as  to  make  the  intersections  of  the  true  poly- 
gon and  curve  well  defined. 


ARCHED   RIB    WITH  FREE  ENDS. 


247 


The  effect  of  such  an  exaggeration  may  easily  be  shown 
since  the  various  steps  of  the  construction  remain  precisely 
the  same.  Suppose  AC  to  be  m  times  as  large  as  it  would  be 
made  by  the  scale  according  to  which  the  span  BD  is  laid  off; 
w  denotes  the  degree  of  exaggeration.  The  distance  /  will  be 
m  times  as  great  as  it  ought  to  be,  and  consequently  the 
height  of  the  equilibrium  polygon  will  be  increased  beyond 
its  true  value  in  the  same  ratio.     But  if  the  true  height  is 

only  —  of  that  found,  the  true  pole  distance  will  be  ;«  x  CO, 
in 

and  LC  must  be  made  equal  to  that. 

The  span  has  been  supposed  half  covered  by  the  moving 

load,  but  any  other  portion  might  have  been  taken  as  well. 

It  is  to  be  noticed  that  the  method  is  perfectly  general,  and 

entirely  independent  of  the  character  of  the  curve  BAD,  or 

of  the  loading. 

Art.  49.— Arched  Rib  with  Free  Ends. 

The  treatment  of  the  arched  rib  with  free  ends  is  not  dif- 
ferent in  any  respect,  except  one,  from  that  given  in  the  pre- 
vious case.  The  exception  is  this,  that  the  condition  2  M  =0 
must  be  omitted,  since  the  bending  moments  at  the  free  ends 
must  disappear. 

In  this  case,  again,  the  centre  line  is  divided  into  equal 
parts.  Observations  made  under  this  head  in  the  preceding 
Article  apply  here  also. 

The  expression  '2nP '  =  2  -^,  sometimes  called  the  bend- 

ing,  is  of  course  based  on  the  common  theory  of  flexure,  and 
denotes  simply  the  difference  of  inclination  of  the  neutral 
surface  of  a  straight  beam  at  the  two  sections  indicated  by 
the  limits  of  the  summation ;    with    a  constant    moment  of 

inertia  it  is  usually  written  j^-  j Mdx.  The  condition  of  fix- 
edness of  the  two  ends  of  the  ribs  requires  the  position  of  the 
neutral  surface  to  remain  unchanged  at  those  two  sections, 
consequently  2nP'   must   be    equal  to   zero   between   those 


248  ARCHED  RIBS. 

limits.  In  the  case  of  free  ends,  the  position  of  the  neutral 
surface  may  be  any  whatever  consistent  with  the  elastic  prop- 
erties of  the  material  at  those  sections.  The  middle  points 
of  the  free-end  sections  must,  however,  retain  their  primitive 
positions,  or  the  summation  of  the  deflections,  either  horizon- 
tal or  vertical,  between  those  points  must  be  equal  to  zero. 
The  only  remaining  conditions,  therefore,  are  ^^^nMy  =  o  = 

But  the  ends  of  the  rib  may  have  any  relative  vertical 
movements,  whatever,  without  changing  the  circumstances  of 
bending.  Consequently  the  only  condition  to  be  fulfilled  is, 
'2.  j^nMy—  o. 

The  same  amount  and  proportion  of  loading  will  be  taken 
as  in  the  previous  case;  the  same  radius,  span,  notation,  and 
scale  will  also  be  taken.  The  figure  of  PL  VI.  represents  the 
construction.  The  moving  load  is  assumed  to  cover  the  half 
span  BC. 

As  before,  take  C  as  the  pole  for  the  trial  polygon,  then 
make  B  5  equal  to  a  half  panel  fixed  load,  supposed  applied 
at  yJ,  and  ^5  6  a  half  panel  (fixed  +  moving)  load,  supposed 
applied  at  the  same  point  ;  also  make  5—4  equal  to  load  at 
b^,  and  6  —  7  the  load  at  b"^^ ,  etc. 

Draw  radial  lines  from  C  to  the  points  of  division,  i,  2,  3, 
4,  5  •  •  •  •  10,  and  construct  the  polygon  E,  a-^,  a2,  a^,  .  .  .  . 
F,  precisely  as  before ;  in  truth  it  is  exactly  the  same  polygon 
that  was  used  in  the  preceding  case.  Since  there  can  be  no 
bending  moments  at  B  and  D,  the  equilibrium  polygon  must 
pass  through  those  points,  hence  EF  is  the  closing  line  of  the 
trial  polygon,  ECF. 

As  has  been  seen,  the  only  condition  to  which  the  equilib- 
rium polygon  is  subject  is  2^  nAIy  =  o  ;  or,  as  before,  as  n  is 
constant. 

:ElMy  =  ^iMay  -^^M.y  =  O. 
Or,  2lA/,y=:Eyf,y. 

The  method  of  constructing  the  deflection  polygon  is  pre- 


ARCHED   RIB    WITH  FREE  ENDS.  249 

cisely  the  same  as  that  followed  in  the  previous  case,  only  in 
the  present  one  a  half  of  the  ordinates  representing  M^  and 
J/ft  will  be  laid  off  from  A  in  order  to  keep  all  the  points  s 
and  t  within  the  limits  of  the  diagram. 

As  the  half  intercepts  at  the  distance  -  from  B  and  E  are 

very  small,  and  as  their  omission  will  lead  to  simplicity  in  the 
diagram,  and  not  cause  much  of  an  error,  they  will  be  neg- 
lected. In  an  actual  case,  however,  the  omission  should  be 
made  with  caution. 

Since  the  moments  at  B  and  D  are  zero,  make  As  equal  to 
\cxb\  ss  equal  to  \c.^b" ,  s's'  equal  to  \cjj"\  s"s"  equal  to  )^cj)'"' ; 
then  draw  horizontal  lines  through  d'd",  b"b"',  etc.,  cutting 
AC  2,%  before.  As  the  end  moments  are  zero,  d'  will  be  on 
AC,  then  d'd"  will  be  parallel  to  Cs,  d"d"'  will  be  parallel  to 
Cs',  and  so  on  until  the  point  d^'  on  the  line  RS  is  reached. 

The  two  portions  of  the  deflection  polygons  for  the  mo- 
ments Ma  will  not  be  similar,  yet  it  is  only  necessary  to  con- 
struct a  single  deflection  polygon,  as  was  shown  in  the  pre- 
ceding Article.  The  sums  .^"^  J/^j'  and  ^^  J/^j'  may  each  be 
divided  into  pairs  of  terms,  each  member  of  the  pair  having 
the  same  value  oi y  ;  this  may  be  done  for  any  case  in  which 
the  moments  may  be  taken  in  pairs.  The  moments  in  each 
pair  of  terms  will  of  course  be  located  equidistant  from  AC. 
Make,  therefore.  At  equal  to  \  (a^d  +  rt'/*),  /'/,  =  to  h  {^-d'  + 
^8<^^"'),  t,t„  equal  to  hagc'"  +  a-^c"^^),  and  tj„,  equal  to  ^  {a^c^  + 
a^c"^^).  Draw  radial  lines  as  shown,  and  make  d/  d/'  parallel 
to  Ct,  d/'d/"  parallel  to  Ct,,  etc.,  until  the  point  ^/  is  reached. 

Lay  off  on  /iC  produced,  CM  equal  to  Ad,'^,  draw  J/A^  par- 
allel to  BD,  and  with  a  radius  CV  equal  to  2{Ad^)  find  the 
point  TV,  and  produce  CN  through  that  point.  In  order  to 
find  the  true  equilibrium  polygon  it  is  only  necessary  to  in- 
crease the  ordinates  of  the  trial  polygon  in  the  ratio  of  CM 
to  CN. 

Hence  draw  d'p  parallel  to  CD,  then  make  Ca^  equal  to  Cp ; 
d'  will  be  one  point  in  the  true  equilibrium  polygon.  In  the 
same  manner  make  Cg  equal  a-j/c'",  and  determine/'  as  before, 


2  CO  ARCHED  RIBS. 

then  make  c^'"  equal  to  Cp' ;  o!"  will  be  another  point  in  the 
true  polygon.  A  shorter  way  to  proceed,  however,  is  the  one 
indicated  in  the  previous  case.  Draw  CL  parallel  to  EF,  then 
a  line  parallel  to  BC  through  L.  Make  CP  equal  to  BC,  and 
draw  OP  parallel  to  BD  ;  OC  is  the  true  pole  distance.  Make, 
therefore,  LC  equal  to  OC,  and  6"  will  be  the  true  pole  from 
which  radial  lines  are  to  be  drawn  to  i,  2,  3,  4,  5,  .  .  .  .    lO. 

Starting  from  any  given  point,  as  B,  make  Ba'  parallel  to 
C  10,  a'a"  parallel  to  6"' 9,  etc.,  etc. ;  the  side  parallel  to  C  i 
should  pass  through  the  point  D. 

The  two  methods  should  be  used  to  check  each  other,  as 
was  indicated  in  the  previous  case. 

Since  LC  is  3*  inches,  TJ^  —  3.^  x  15  =  52.5  tons,  and  its 
line  of  action  is  evidently  BD. 

A II  those  directions  of  a  general  character  zvhich  accompany 
and  follow  the  construction  in  the  preceding  case  apply  with 
equal  force  to  the  present  one  and  those  which  follow.  It  is 
particularly  important  in  the  graphical  treatment  of  all  arched 
ribs  to  make  the  polygons  approach  as  nearly  their  ultimate 
limits,  i.  e.,  curves,  as  possible  ;  for  that  reason  it  will  be 
advisable  in  most  cases  to  divide  the  actual  panels  into  two 
or  more  equal  parts  in  the  search  for  the  true  equilibrium 
polygon. 

Art.  50. — Thermal  Stresses  in  the  Arched  Rib  with  Ends  Fixed.* 

Thermal  stresses  are  those  stresses  which  are  co-existent 
with  any  variation  of  temperature,  in  the  structure  consid- 

*  In  reality  the  deflection  which  produces  stress,  in  the  case  of  variation  of 
temperature,  is  not  the  whole  deflection.  If  the  points  B  and  Z>,  in  PI.  VII., 
were  free  to  move,  there  would  be  no  thermal  stresses,  but  there  would  be  deflec- 
tion. This  deflection,  in  the  present  case,  would  be  (if  loo  units  become  100.12204 
for  an  increase  of  180"  F.) 

(100  -i — -—0.12204  I  -^  100  X   19.1  —  19.1  =  .021  ft. 

Strictly  speaking,  the  deflection  to  be  used,  then,  for  a  change  of  165°  F.  is 
0.25  —  0.021  =  0.23  ft.  The  difference  is  so  small,  however,  that  it  may  be 
neglected,  especially  since  the  error  is  a  small  one  on  the  side  of  safety.  These 
observations  are  general  and  apply  to  all  cases. 


THERMAL    STRESSES  IN  ARCHED  RIB.  25  I 

ered,  and  whose  values  depend  upon  that  variation.  Any 
variation  of  temperature  in  the  material  of  which  an  arched 
rib  is  composed  will  cause  a  variation  in  its  length,  and  con- 
sequently a  deflection  at  any  given  point.  Although  the 
temperature  is  supposed  to  change,  yet  the  ends  of  the  rib 
are  supposed  to  remain  in  their  normal  positions,  so  that  the 
general  conditions  ^>/J/=:oand  ^)iMy  —  o  =  .^'«J/;ir,  hold 
for  thermal  stresses  as  well  as  for  any  other,  remembering 
that  11  is  constant. 

Any  change  of  form,  such  as  that  arising  from  the  applica- 
tion of  loading,  will  cause  extra  stresses,  which  are  to  be  de- 
termined in  precisely  the  same  manner  as  that  used  for  ther- 
mal stresses  ;  in  fact,  they  belong  to  the  same  class  of  stresses. 

Every  kind  of  material  has  its  own  coefficient  of  linear  ex- 
pansion ;  wrought  iron,  for  instance,  expanding  .12204  units 
in  every  100  for  a  change  of  temperature  from  32°  to  212°  F., 
while  tempered  steel  gives  the  empirical  quantity .  1 2396  for  the 
same  conditions.      (D.  K.  Clark,  "  Rules,  Tables,  and  Data.") 

The  arched  rib  shown  in  PI.  V.,  and  already  treated  for 
ordinary  stresses,  will  be  supposed  to  be  of  such  a  material 
and  to  be  subjected  to  such  a  change  of  temperature  that  the 
point  A  will  suffer  a  vertical  deflection  of  3  inches.  It  is  a 
matter  of  indifference  in  which  direction  the  deflection  takes 
place ;  it  will  be  supposed  upward  in  the  present  case. 

The  effect  of  the  thermal  variation  is  to  cause  bending 
moments  at  the  various  sections  of  the  rib  from  which  the 
deflection  results.  Also,  to  keep  the  ends  in  their  original 
positions  requires  the  existence  of  a  horizontal  force,  such  as 
the  stress  which  may  be  supposed  to  exist  in  a  horizontal  tie. 
The  stresses  and  bending,  then,  caused  by  thermal  variations 
are  the  same  as  those  which  would  be  caused  by  a  horizontal 
force  having  the  proper  line  of  action  ;  the  problem  then 
resolves  itself  into  finding  the  proper  value  and  line  of  action 
of  this  horizontal  force. 

The  figure  to  be  used,  and  which  will  be  referred  to,  is  that 
of  PI.  VII.,  and  represents  the  same  rib  precisely  as  PI.  \'^. 
For  the  sake  of  greater  accuracy, ;/  will  be  taken  half  as  great 
as  in  Pis.  V.  and  VI. ;  its  value  will  then  be  5.47  feet. 


252 


ARCHED   RIBS. 


It  will  first  be  assumed  that  the  points  b,  b^,  b^,  etc.,  are  at 
a  uniform  horizontal  distance  apart  of  5  feet,  and  at  the  same 
time  equidistant  on  the  curve ;  that  which  might  be  allowable 
in  a  very  flat  curve. 

The  conditions  '2^  nM  =  o  and  2^?iMx  —  o  show  H" K" 
to  be  the  line  of  action  of  the  horizontal  force,  which  will  be 
called  T^.  The  line  H"  K"  is  the  same  line  as  N"K"  in  PI. 
v.,  since  it  is  located  by  exactly  the  same  condition,  the 
vertical  intercepts  between  it  and  the  curve  BAD  represent- 
ing the  moments  M.  Since  the  only  force  acting  on  the  rib 
is  7)i,  the  bending  moment  M  will  be  equal  to  T,,  multiplied 
by  the  proper  vertical  intercept.  Thus  the  moment  at  b,  will 
be  T^  X  a^b/,  and  that  at  bg,  T^  x  a^bg. 

The  value  of  Tf^  is  determined  by  either  of  the  conditions 
2^n3fy=nf,E/,  or  2^^nMx=I)£/;  by  way  of  variety  the  lat- 
ter will  be  taken.  Assume  any  point  as  C  for  the  pole,  and 
make  ^(T  the  pole  distance.  Makei5i  equal  to  i^^^g;  i  —2  equal 
to  agbg ;  2  —  3  equal  to  a^b^ ;  .  .  .  .  and  9— 10  equal  to  ab,  and 
construct  the  polygon  Caa" .  .  .  .  a^^  in  the  usual  way.  B  —  10 
is  one  half  //'b',  and  a^N'  is  a  vertical  line  through  //,  while 
Bb'  is  one  quarter  of  Bb.  Draw  the  horizontal  line  a^E,  then 
the  vertical  intercepts  included  between  a^£  and  the  deflec- 
tion polygon  a^d^C,  of  the  kind  iaJi),  represents  in  an  exag- 
gerated manner  the  deflections  (vertical)  of  the  points  in  the 
rib  vertically  above  them. 

Now  since  the  actual  moment  in  an  equilibrium  polygon  is 
equal  to  the  vertical  ordinate  multiplied  by  the  pole  distance, 
the  actual  vertical  deflection  at  A  is  CE  multiplied  by  its  pole 
distance ;  however,  since  the  real  deflections  are  proportional 
to  the  vertical  ordinates  of  the  kind  [aJi),  those  deflections 
may  be  at  once  -found  by  multiplying  those  ordinates  by  a 
proper  ratio.  That  ratio  is  a  known  quantity,  because  the 
real  deflection  at  A  is  known  to  be  three  inches.  The  ordi- 
nate CE  measures  1.61  inches  on  the  drawing,  and  represents 
16. 1   feet  on  the  actual  rib;    the    ratio    desired  is  therefore 

^     =  — - —  =  ^ .     To  illustrate,  the  deflection  at  the 

12  X  16.1       193.2      64.4 


THERMAL    STRESSES  IN  ARCHED  RIB. 


253 


^section  /  is  13.9  -^-  64.4  =  0.2 16  feet.  Similarly,  the  deflection 
at  ^3  is  6.2  ^  64.4  =  0.096  feet.  These  quantities  will  be  used 
farther  on. 

It  will  now  be  necessary  to  give  a  little  consideration  to  the 
general  equation  '^hjSIx  —  DEI. 

If  the  ordinates  of  the  kind  {ah),  measured  from  H" K'\  are 
denoted  by  y\  M,  from  what  has  already  been  said,  can  be 
written  as  T^^y  .     Hence  the  general  equation  may  be  written, 

2nMx  =  T^^ny'x  =  DEI. 

Or,  since  n  is  constant, 


n  2j/  x ' 

This  last  is  the  equation  from  which  7\  is  to  be  found. 

Now  since  j'  is  positive  or  negative  according  as  it  is  meas- 
ured on  one  side  or  other  of  the  line  H" K" ,  and  since  n  is 
assumed  to  be  uniform  and  horizontal,  the  quantity  n'^y  x  is 
the  difference  between  the  statical  moments  of  the  moment 
areas  on  the  different  sides  of  that  line  in  reference  to  the  sec- 
tion considered.  Written  as  an  integral  expression,  it  would 
be  fy'xdx. 

The  area  of  the  surface  Ab(,ina^  is  1.12  sq.  in.,  or  112  sq.  ft. 
full  size.  The  distance  of  the  centre  of  gravity  of  the  same 
area  from  AC  is  very  simply  found  by  construction.  Take 
any  point  «,  for  a  pole,  and  a-jm  for  the  pole  distance.  Make 
rniH]^  =i?im^=  ^a-ib-i,  nixiUi  equal  to  a(,b^,  and  so  on,  making, 
however,  nif^nii  equal  to  a  half  of  Aa^,.  Then  construct  the 
equilibrium  polygon  r,  e-^,  e^,  .  .  .  .  c-.  The  sides  ee^  and  e^e-f 
produced  will  cut  each  other  in  the  vertical  line  GH  passing 
through  the  centre  of  gravity  of  the  area  Ama^. 

The  area  BH  in  is  to  be  treated  in  precisely  the  same  man- 
ner, taking  a^  as  pole,  and  a^H  '  as  pole  distance,  and  making 
Bn  equal  to  a  half  of  BH" .  The  vertical  line  KL  passing 
through  the  centre  of  gravity  of  the  area  is  found  as  before  at 
the  intersection  of  the  sides  ^i  and/s/g  prolonged. 

The  area  BH'ni  is,  of  course,  equal  to  the  area  Ama^,  con- 


2  54  ARCHED   RIBS. 

sequently  the  value  of  7i'2, y'x  for  the  section  A  will  be  the 
product  of  the  common  area  by  the  horizontal  distance  be- 
tween their  centres  of  gravity,  i.  e.,  3.26  inches  in  the  figure, 
but  32.6  ft.  full  size. 

The  cross-section  of  the  rib  will  be  assumed  to  be  of  such 
form  that  £/has  the  value  of  2,000,000  foot-tons.  Hence  for 
the  point  .'^,Z>=  3  inches  or  0.25  ft.,  7t'2y'x  =  112x32.6  = 
3651.2; 

^        2000000  X  0.25 

•■•    ^A= z ^  =  137  tons. 

3651.2  -^^ 

For  any  other  section,  as  /,  the  deflection  is  13.9  -^  64.4  == 
0.216  ft.  The  vertical  line  passing  through  the  centre  of 
gravity  of  the  area  included  between  ;//  and  a  vertical  line 
through  /  passes  through  the  intersection  of  the  sides  e^a-j  and 
ec^,  prolonged,  and  the  distance  between  KL  and  it  is,  as 
shown,  3.04  inches  in  the  drawing.      For  this  section 

n2yx  =  1 12  X  30.4  —  49.82  X  5.8  =  31 16; 

™        2000000  X  0.216 
•■■   ^•= ^776 ='39  tons. 

In  precisely  the  same  manner,  for  the  point  h^, 

™        2000000  X  0.C96 

7fc  = ^—  =  nS  tons. 

1421  -^^ 

The  values  should  have  been  the  same,  except  for  the 
errors  incident  to  a  small  scale  and  the  fact  that  polygons 
were  used  where  curves  really  belonged  ;  yet  the  difference 
between  the  extreme  values  is  only  2^  per  cent,  of  the  larger, 
which  is  not  very  much  of  an  error. 

If,  however,  the  true  value  of  n  (5.47  feet,  nearly)  be  taken 
along  the  centre  line  of  the  rib,  a  decidedly  different  result  will 
be  found,  since  n'2 y'x  will  have  a  different  value. 

The  quantity  x,  as  before,  will  be  measured  from  a^  to  the 
intersections  of  the  dotted  lines  drawn  through  the  points  {b), 
while  j'  will  represent  any  vertical  ordinate  (belonging  to  any 


THERMAL    STRESSES  IN  ARCHED  RIB. 


25c 


point  {bf)  from  the  line  H" K",    taken  positive   downward. 
The  point  m  will  be  assumed  midway  between  a^  and  a^ . 

The  following  values  are  measured  from  the  original  draw- 
ing: 

=  —  6.2       I  {xy)  =  —     4-22 
=  -6.1  xy'  =  -    33-25 

=  —  5.5  xy'  =  —    60.00 

=  -  4.6  xy'  =  -    74.52 

=  -  3.2  xy'  =  -    69.44 

=  -  1-35  ^J  =  -    36.18 


x=     1.36 

j' 

x=    5.45 

7 

.r  =  10.9 

y 

x=  16.2 

y 

jr  =  21.7 

y 

X  =  26.8 

y 

- 

-  277.61 

X  =  32.0 

y=  0.7 

;rj/  = 

22.4 

X  =  36.8 

y=  3-2 

;rj'  = 

117.76 

X  =  41.6 

j'  =  6.0 

xy  = 

249.6 

A-  =  46.0 

y  =    9.2 

.rjj/  = 

423.2 

X  —  49.0 

j'  =  11.98 

n^j)  = 

293-5 

1084.06 

The  first  values  of  x  and  y'  belong  to  that  portion  of  the 
moment  area  adjacent  to  Aac,. 

The  last  values  of  x  and  y'  belong  to  that  portion  of  the 
moment  area  adjacent  to  BH  '  \  and  half  of  each  product  is 
taken,  so  that  it  may  be  multiplied  by  the  full  value  of  n  = 
5.47  feet. 

The  formula  then  gives: 

„        DEI         0.25  X  2000000  ^        ,         1  s  « 

7\  =  — ^-^— -  = =  I  n-3  tons  (nearly).* 


y  X 


4411.30 


*  The  method  by  deflection  polygon  given  in  the  next  Article,  produces  a  re- 
sult essentially  the  same  as  the  one  above. 

£C  is  the  pole  distance  laid  down  to  a  scale  of  400,000  foot-tons  to  the  inch. 


n  = 


4  X   16. 1   X  5.47 


0.62  X  400000 
6^2 


—  113.6  tons. 


The  agreement  is  much  closer  than  can  ordinarily  be  expected  with  the  scales 
used. 


256 


ARCHED   RIBS. 


The  difference  between  the  results  of  the  two  methods  is 
137.00  -  113-3  =  237  tons. 

This  difference  is  by  no  means  small,  and  shows  how  care- 
fully the  approximate  method  ought  to  be  used. 

With  a  wrought-iron  rib,  the  deflection  taken,  3  inches  at 
the  middle  of  the  span,  belongs  to  a  change  of  temperature 
of  about  165°  F.,  a  very  extreme  case,  which  accounts  for  the 
large  values  of  Tj^. 

This  shows,  however,  in  a  very  marked  manner,  the  impor- 
tance of  putting  together  an  arched  rib  at  about  the  mean 
temperature,  for  then  the  variation  of  temperature  to  be  taken 
in  the  calculation  of  thermal  stresses  will  only  be  about  half 
the  variation  between  the  extreme  limits. 

The  effect  of  T^  is  the  same  as  if  that  force  were  applied  at 
m  and  acting  toward  H"  for  the  portion  Bm  fixed  at  B,  but 
applied  at  m  and  acting  toward  vi  for  the  free-end  portion 
(so  considered)  mA. 

The  change  of  temperature,  165°,  changes  the  radius  from 
75  feet  to  74.536  feet,  and  increases  the  length  of  the  curve 
0.122  feet. 

Art.  51. — Thermal  Stresses  in  the  Arched  Rib  with  Ends  Free. 

The  method  to  be  used  in  the  present  case  is  somewhat 
shorter  and  simpler  than  the  one  used  in  the  preceding,  but 
will  probably  not  give  as  nearly  correct  results  when  the  scale 
used  is  small. 

The  figure  to  be  used  is  that  shown  in  PI.  VIII.,  and  the 
curve  BAD  is  precisely  the  same  as  that  shown  in  Pis.  V., 
VI.,  and  VII. 

For  the  sake  of  greater  accuracy,  the  curve  BA  will  be  di- 
vided into  ten  equal  parts.  There  will  then  result,  n  =  bby  = 
bxb%  =  etc.,  =  5.47  (nearly)  feet. 

Since  BD  is  the  true  position  of  the  closing  line  for  free 
ends,  the  effect  of  the  variation  in  the  temperature  will  be 
the  same  as  that  of  a  horizontal  stress,  T^  ,  whose  line  of 
action  is  BD,  and  which  will  produce  a  deflection  equal  to 
the  thermal   deflection.     What  maybe  called  the  "thermal 


THERMAL    STRESSES  TV  ARCHED   RIB.  257 

moment,"  therefore,  at  any  point  will  be  equal  to  7\  multi- 
plied by  the  vertical  ordinate  of  the  curve  at  that  point,  or 
M—  Tf^y'.     The  moment  at  b^,  for  instance,  is  M  =  T^  >^b^n^. 

Supposing  the  rib  to  be  of  wrought  iron,  a  change  of  tem- 
perature of  137°  F.  will  cause  the  length  to  change  from 
109.454  feet  to  109.556  feet,  and  the  radius  from  75  feet  to 
74.67  feet ;  the  corresponding  upward  vertical  deflection  at 
crown  A  will  be  i.^  inches,  or  ^  of  a  foot. 

Let  the  two  general  equations  be  compared : 

2Px      =  T^y. 
2nMx  =  2M'x  =  £1.1). 

From  these  two  equations  it  is  seen  that  if  M'  be  taken  as 
vertical  loading,  and  £1  as  pole  distance,  then  the  ordinates 
of  the  resulting  equilibrium  polygon  will  represent  the  deilec- 
tions  according  to  the  same  scale  by  whicJi  x  is  measured.  It  is 
important  also  to  notice  that  M'  and  £1  are  of  the  same  de- 
nomination (foot-pounds  or  foot-tons,  as  the  case  may  be), 
consequently  M'  is  to  be  measured  in  the  same  scale  as  that  ac- 
cording to  which  £1  is  laid  dozvn. 

Since  nM  =  n  Tj,y'  {n  being  constant),  the  vertical  ordinates 
of  the  kind  ((^«)  are  proportional  to  the  moments  J/ or  J/', 
and  they  may  be  taken  to  represent  those  moments  ;  since, 
however,  that  would  carry  the  lower  extremity  of  the  load 
line  B  —  10  off  the  diagram,  one  third  only  of  those  ordinates 
will  be  taken  in  the  plate. 

As  before,  £/  will  be  assumed  to  be  2,000,000  foot-tons, 
and  the  scale  according  to  which  it  is  to  be  laid  down  for  the 
pole  distance  at  200,000  foot-tons  to  the  inch.  Hence  make 
£7//^  parallel  and  equal  to  BCD.  Since  BD  is  ten  inches  in 
the  diagram,  ££  is  the  pole  distance,  and  £  will  be  taken  as 
the  pole. 

As  in  the  case  of  the  same   rib  with  external  loading,  the 

half  intercept  at  the  distance  -  from  B  will  be  neglected. 

4 
Make  £1  equal  to    ^{\AC),    i-2  equal    to   ^b^ns,  etc.,  and 
construct  the  polygon  a  a^ai //"  in  the  usual  man- 

'7 


258 


ARCHED   RIBS. 


ner.  BA  is  divided  into  ten  equal  parts  for  the  sake  of  greater 
accuracy. 

The  polygon  thus  constructed  will  represent  the  actual  de- 
flections to  a  scale  of  lo  feet  to  the  inch. 

Now  Ea  is  equal  to  1.12  (3.32  inches  on  the  original  draw- 
ing) inches,  or  11.2  feet  full  size,  whereas  it  ought  to  be  but 
0.125  foot  ;  and  since  the  pole  distance  is  to  remain  the  same, 
the  moments  must  be  reduced  in  the  ratio  of  \  to  13.2  ;  or  the 

moments   as   actually  taken  must  be  multiplied  by  - —         = 

^  ^  8  X  II. 2 

^.     The   reduction  for  the   bending   moment   at  ^-J,  there- 

r         .    ,     .  ^  „    ^       0.636666  X  200000  .  ^ 

fore,  IS  \  AC  y.  200000  -=-  8q.6  =  — ^ ^r — ^ ,  smce  AL 

"*  ^  89.6 

is  equal  to  1.91  inches.      Hence, 

„,       M'      0.636666x200000      ^      ,^ 
Tn  =  — -  =  — ^-TT^ =  7446  tons. 

*      y  89.6x19.1 

But  since  Af  =  ?zM: 

Ti,  =  —  —  —  -  =  —     =  13.6  tons. 
y         ny  n 

This  operation  may  be  considerably  shortened  by  remem- 
bering that  0.636666 -^  19.1  =  I  -^  30,  and  that  this  ratio  is 
constant  for  all  points.  If,  therefore,  the  moment  at  any 
other  point,  as  b^,  be  taken,  precisely  the  same  result  will  be 
obtained. 

The  method  used  in  the  preceding  Article  gives,  at  least 
approximately,  the  same  result.  Taking  Bn,  Bn^,  etc.,  for 
the  different  values  of  x,  and  bn,  b^n^,  etc.,  for  r,  there  will 
result  : 

X  =    4.0  feet,  y'  =    3.7  feet,  .-.  xy'  =     14.8 

X  =    8.4     "  y'  =    6.8    "  .-.  xy'  —    49.6 

X  =  13.2     "  y  —    9-5     ''  •'■  'V  =  1 01.6 

x=  18.0     "  y  =  12.0    "  .-.  xy'  =  183.6 


ARCHED   KfBS    WITH  FIXED   ENDS. 


259 


X  =  23.1  feet,  }>'  =  14.2  feet, 

X  =  28.2    "  y  =  16.0  " 

x=  33.6    "  y  =  17.4  " 

X  =  39.0  "  y  =  18.3  " 

X  —  44.4   "  y  =  18.9  " 


^y  =  307-2 

xy  =z  425.8 

-I'j'  =  554-4 

xy  =  643.5 
xy  =  799.2 


With  these  values  the  partial  summation  is  : 
2:^iyx  =  n^j'x  =  16846.00. 

A  product  for  the  moments  adjacent  to  B  is  nearly  : 

0.92  X    I   X  2.73  =  2.51. 
Another  for  those  adjacent  to  AC  is  nearly  : 
19.1  X  48.64  X  2.73  =  2536.5. 

Taking  the  sum  of  these  results  for  the  complete  summa- 
tion : 

„         DEI        0.125  X  2000000 

7*  =    s:^ — -  =  5 =13.0  tons  (near  y). 

"        2.nrx  19385.00  -^  \         Jj 

The  polygon  a,  a^,  a^^,  a^ //  is  an  exaggerated  repre- 
sentation of  the  movements  of  the  points  b,  b^,  b^,  etc.,  when 
the  temperature  is  changed  137°  F. 

In  all  cases,  as  large  a  diagram  as  possible  must  be  used,  in 
order  to  reduce  the  scale  for  the  pole  distance,  so  that  that 
distance  may  be  the  largest  possible. 

The  use  to  be  made  of  T^  will  be  shown  farther  on. 

Art.  52. — Arched  Rib  with  Fixed  Ends — /  and  n  Variable. 

The  rib  to  be  taken  in  this  case,  and  the  loading,  are  pre- 
cisely the  same  as  taken  in  the  preceding  cases.  As  before, 
the  centre  line  is  divided  into  ten  equal  parts  of  10.95  feet 
each.     The  data  are  therefore  the  following : 

Span  =  100.00  feet. 

Radius  =     75.00     " 

Panel  fixed  load       =      4.00  tons. 
Panel  moving  load  =     10.00     " 
Centre  rise  of  rib     =     19.10  feet. 


26o  ARCHED  RIBS.^ 

The  moving  load  is  supposed  to  cover  the  half  span  BC. 

The  figure  to  be  referred  to  is  that  shown, in  Pi.  IX. 

The  point  h  is  midway  between  b'  and  b'\  while  k  is  mid-^ 
way  between  b"  and  b" . 

The  moment  of  inertia  /of  the  cross-section  of  the  rib  will 
be  taken  as  2,000,000  foot-tons  throughout  Bh,  1,777,778  foot- 
tons  throughout  hk,  and  1,600,000  foot-tons  throughout  kA. 
The  same  values  hold  for  similar  portions  of  AD. 

I,  2,  3,  etc.,  ....  10,  is  the  load  line,  and  ^C  the  pole  distance 

of  the  equilibrium  polygon  Ea^  Ca-fF.     That  polygon  is  drawn 

in  precisely  the  same  manner,  in  fact,  is  precisely  the  same 

one  as  that  shown  in  PI.  V. ;  the  radiating  lines  drawn  from 

C  are,  therefore,  omitted.     As  the  ends  are  fixed,  EF  cannot 

B  nM 
be  the  closing  line;  but  the  condition  ^     — —  =  o  must  first 

be  imposed. 

It  has  already  been  shown  that  for  any  point  in  the  arched 
rib  the  moment  M  =  M^  —  M^. 

Hence  if /j  be  the  moment  of  inertia  for  the  portion  2Akoi 
the  rib,  there  may  be  written : 

For  Bh,  i  has  the  value  j  =  0.8  ;  for  M,  the  value  ^  =  0.9  ; 

and  ior  Ak,  the  value  -  =  i.o. 
-'I 

The  closing  line  HK  must  be  so  located  that : 

But  for  the  equilibrium  polygon  Ea-^a^,  etc.,  the  summation 
'2  ^  niMa  has  the  value  : 

2  j^  niMa  =  O.Sn'  x  vv'  +  o.8n  x  a^c'  +  O.gn  x  rto<r''  —  n  x 
c"'a^  —  ;/  X  ^^4  —  n  X  c^C  —  n  x  ^^V?6  —  n  x  f^a,  —  0.9«  x 
c^^a^  +  o.2,n  x  a^c"^  +  o.8«'  x  v"'v"=o. 


ARCHED  RIB    WITH  FIXED  EXDS.  26 1 

On  the  curve  BAD  the  distance  Bb  is,  -  =  \^  Bb' ,  and  H'v 

4 
is  taken  vertically  through  b\    also  Db'^  —  Bb    and   K'v"'  is 
drawn  vertically  through  b'^.     From  the  location  of  b  and  b'^  it 

follows  that  n'  =  — ,  consequently  ?i  may  be  canceled  from  the 

vv'  l>"'l/' 

series  by  writing  —  and  instead  of  the  whole  quantities 

22 

themselves. 

If  vv  be  taken  at  1.04  inches  and  v"'v"  at  0.66  inches,  the 
above  summation  (after  dropping  11  in  the  manner  shown) 
gives  a  result  of  —  0.02  of  an  inch  only.  This  agreement  is 
sufficiently  close. 

The  vertical  intercepts  and  their  products  by  /,  are  the  fol- 
lowing : 

+  0.8  X  —    =  +  0.8  X  --^  =  +  0.42 
2  2 

+  0.8  X  a^c'    =  +  0.8  X  0.56  =  +  0.45 

+  0.9  X  a./'  =  +  0.9  X  0.06  =  +  0.05 

—  I  X  a^c'"  =  —  I  X  0.30  =  —  0.30 

—  I  X  a^c^"  =  —  I  X  0.50  =  —  0.50 

—  I  X  Cc"'    =  —  I  X  0.46  =  —  0.46 

—  I  X  a^c"^  =  —  I  X  0.31  =  —  0.31 

—  I  X  rt-^r^'  =  —  I  X  0. 10  =  —  0. 10 

+   0.9    X  ^gC^'^zz:    +   0.9    X    0.15    =    +0.13 

+  0.8  X  rtgc^  =  +  0.8  X  0.42  =  +  0.34 

o  V'v"'  _  0.66  ^ 

+  0.8  X =  +  0.8  X  =  +  0.26 

2  2 

The  other  condition  for  the  closing  line  HK'\s  2   nMatx  =  O. 


262  ARCHED  RIBS. 

Taking  the  products  of  the  ordinates  (nMi)  of  different  signs, 
by  their  horizontal  distances  from  DK",  as  was  done  in  Art. 
48,  there  will  result  the  sums  +  91.8  and  —90.5.  The  alge- 
braic sum  is  only  +  1.3,  which  is  near  enough  to  zero.  HK 
will  therefore  be  taken  as  the  proper  closing  line. 

The  condition  which  locates  the  closing  line  H"K''  is 
similar  to  that  which  placed  HK\  it  is  the  following: 

2   tiiMi,  =  0.8  X  2n'  y.  H'b  -\-  0.8  x  2n  x  c^b'  +  0.9  x  2n  Y-c^b' 
—  2n  X  b"'c^  —  2n  x  b^c^  —  fi  x  Acr,  =  o. 

Since  the  curve  BAD  is  symmetrical  in  reference  to  AC, 
the  line  H"K"  will  evidently  be  horizontal.  For  the  same 
reason,  2n'  and  2n  are  written  in  the  summation  in  all  its 
terms  except  the  last.  As  before,  Jt  is  one  half  «,  and  by  so 
writing  it,  n  may  be  dropped  from  the  series. 

By  making  Ac^  —  0.58  inch,  the  summation  (after  dropping 
n)  gives  2.18  —  2.16  =  0.02  only;  H"K"  will  therefore  be 
assumed  to  be  the  proper  closing  line  for  BAD. 

The  vertical  intercepts  and  their  products  by  i  are  the  fol- 
lowing : 

o      bH'  o       1. 1 3 

2  X  0.8  X =  2  X  0.8  X —  0.90 

2  2 

2  X  0.8  X  /;>!    =  2  X  0.8  X  0.66  =  1.06 

2  X  0.9  X  b"c.i   —  2  X  0.9  X  0.12  —  0.22 

2  X  b"'c^  =  2  X  0.27  =  0.54 

2   X  //^r4    =  2   X  0.52  =  1.04 

*  Ac5=  0.58  =  0.58 

As  before,  since  the  curve  BAD  is  symmetrical  in  reference 
to  A,  the  two  conditions  -^^/z/J/^  =  o,  and  2: j^niM^x  =  o,  are 
equivalent. 

The  equation  expressing  the  condition  that  the  horizontal 
deflection  of  D  in  reference  to  B  is  nothing,  is  the  general 
one  already  given : 


ARCHED   RIB    WITH  FIXED   EA'DS.  263 

^B  nMy  _    Hi 


Or,  as  in  preceding  cases  : 

KriM^y  =  2lrtM,y. 

The  quantity  n^  is  any  standard  value  of  n,  just  as  /j  is  a 

standard  value  of  /,  and  r  is  such  a  variable  ratio  that  for  any 

n 
section   ^  —  r/i,  or  r  =  —  .     In  the  present  case  n^  will  have 

the  value  10.95  feet ;  consequently  r  will  be  unity  for  all 
sections  except  d,  and  for  that  one  it  will  be  ^. 

The  ratio  r  might  have  been  used  in  the  previous  summa- 
tions of  the  present  Article  in  exactly  the  same  manner  and 
with  exactly  the  same  values  as  in  the  present  one. 

The  ratio  t  has  the  same  value  as  before. 

The  principles  on  which  the  remaining  constructions  are 
based  are  precisely  the  same  as  those  shown  in  the  preceding 
Articles  ;  the  difference  in  the  construction  itself  is  simply 
this,  that  riM^  and  rtil/h  are  taken  instead  of  J/„  and  Mf,.  In 
other  words,  r  and  /,  in  general,  in  this  case,  have  values  dif- 
ferent from  unity,  while  in  the  preceding  cases  r  =  i  —  i. 

The  following  are  the  values  of  {riM^  : 


h. 

X  0.8 

vv     +  v"v")  = 

At', 

X  0.8 

ttxC       +   ^9'-''')     = 

tt'  \ 

X  0.9 

V'  -a^^''')^ 

ft"  ; 

X          [ 

-  a^c"  —  a-f^'''')  — 

-  /"/'" ; 

X         I 

I  X  I  X  CV  == 

-  ft'" ; 

Draw  radial  lines  from  C  to  the  points  / ;  then  draw  the 
horizontal  lines  bC" ,  b'dxd',  b"dx'd\  etc.  C"d^  is  parallel  to 
Ct\  d^di'  is  parallel  to  Ct  \  dx"d^"  is  parallel  to  67",  etc. 
CC'd-^d-l'dx'dx'dx  is  then  the  deflection  polygon  for  the  mo- 
ments M^. 


264  ARCHED  RIBS. 

The  following  are  the  values  for  (riM^ : 

^  X  0.8  X  2H'b        =      As; 
I  X  0.8  X  2b' c^  —      ss' ; 

I  X  0.9  X  2b" C2         —      ss"  ; 
IX      I  X  (-  2b"'c^=  —  s"s"' ; 
IX      IX  (—2/^%)=  —  ^"j-i^; 
IX      IX  {-  Ac^)    =  -  j*M. 

The  point  A  belongs  to  the  curve  BAD. 

The  sides  of  the  deflection  polygon  C"d'd"d"'d^''d^  are 
parallel  to  radiating  lines  drawn  from  C  to  the  points  s. 

Ad^^  represents  ^^j^riM^y,  and  Ad^  represents  ^^j^riM^y. 
Since  the  first  is  less  than  the  second,  the  moments  M^  must 
be  increased  in  the  ratio  of  Ad-^  to  Ad".  Hence  on  AC  pro- 
longed, make  CM  equal  to  Ad-^" ;  draw  T^/iV- parallel  to  BD 
and  with  a  radius  67V  equal  to  Ad"  find  the  point  N.  Pro- 
long the  line  67V;  this  line  will  enable  the  true  moments  Ma 
to  be  determined  in  the  manner  already  shown  in  the  other 
cases. 

Draw  c"p  parallel  to  BD,  c^a"  equal  to  Cp  will  give  a  point 
d'  in  the  true  equilibrium  polygon.  Again,  Ck  "  equals  KF, 
hence  K"a^,  equal  to  Ck'",  gives  the  true  point  a^. 

Also,  6^  is  equal  to  HE  ;  and  H"a,  equal  to  Cp' ,  gives  the 
true  point  a.  All  points  in  the  true  polygon  might  be  thus 
determined,  but  it  is  advisable  to  check  by  the  other  method 
already  shown. 

For  this  purpose  draw  CL  parallel  to  HK,  and  LC  parallel 
to  BD.  Then  make  CP  equal  to  BC  and  draw  OP  parallel  to 
BD.  Take  LC  equal  to  CO.  C  is  the  pole  and  LC  the 
pole  distance  of  the  true  equilibrium  polygon.  Finally  draw 
radiating  lines  from  C  to  the  load  points  i,  2,  3,  4,  5,  6, 
etc.  Starting  from  any  point  already  determined,  as  a,  draw 
ad  parallel  to  C\o;  dd'  parallel  to  C'(^;  d'd'  parallel  to 
C'?>,  etc.  The  different  points  found  by  the  two  methods 
ought  to  coincide. 

The  polygon  add'd"a^"a"d'^a'''^^d'^^^d^a^  is  the  true  equili- 
brium polygon,  which  was  to  be  found. 


ARCHED   RIB    WITH  FIXED  ENDS. 


265 


C L  is  3.86  inches  ;  hence  7"^,  whose  line  of  action  is  H" K" 
is  equal  to  3.86  x  15  =  57.90  tons. 

In  the  determination  of  the  thermal  stresses  the  same  fig- 
ures will  be  used,  and  there  will  be  supposed  such  a  change 
of  temperature  that  the  point  A  will  suffer  a  vertical  deflec- 
tion of  1.5  inches  or  0.125  of  a  foot  ;  the  same,  in  fact,  as  was 
supposed  in  a  previous  case. 

As  the  ends  of  the  rib  are  fixed,  the  general  conditions, 
2j^riM  =  o;  and  2^ri3fj/=2^riAfx  =  o  hold  as  well  for 
thermal  stresses  as  others.  Consequently  H" K"  will  be  the 
line  of  action  of  the  horizontal  stress  7^,  induced  by  the 
variation  of  temperature. 

As  before,  let  y  denote  the  vertical  ordinate  of  any  point 
in  the  curve  BAD  from  the  line  H"K",  then  there  may  be 
written  : 

^1         >r,^         •7,7-  J^  ^h^h      ^A         ■     , 

DEh 


Tn  = 


^A       ■    , 

Hi^.  ^rty  X 


In  order  to  save  confusion  in  the  figure,  ?Zi  will  be  taken  at 

its  previous  value  10.95  feet. 

Also, 

EI^  =  1600000  foot-tons  ; 

D     =  0.125  foot. 
A  half  of  the  moment  at  A  will  be  supposed  applied  at  a 

point  e  distant  —^  from  A  on  the  curve  BAB,  and  none  at 

4 
all  at  A. 

The  co-ordinate  x  will  be  measured  from  c^  towards  K'\ 
The  following  values  then  result: 

X  =    2.72  ft.        riy'  =  ^  X   I      X  5.7  =  2.85       ri'y'x  =      7.752 
x=  10.90  "         riy'  =  I  X  I      x  5.2  =  5.2         riy'x  =    56.68 
X  =  21.7     "         rty'  =1x1      X  2.7  =  2.7         riy'x  =    58.59 

123.022 


266 


ARCHED  RIBS. 


x=   3.2  ft. 
x  =  4i.6  " 
x=4y.g  " 


Hence, 


r?jj''=  —  I  X0.9  X  1.2  =  — 1.08 
r{y=  —  ixo.Sx  6.6  =  — 5.28 
r«y=:— ^xo.8x  1 1.3  =—4.52 


rty,Y  =  -   34.56 
rz)/>  =  —  2 1 9.648 
rtyx  =  — 206. ^oS 


—460.716 


^^rz/x  =  123.022  —  460.716  =  —  337.694. 

The  negative  sign  will  be  dropped  hereafter,  as  it  refers 
simply  to  the  direction  in  which  j  was  measured. 
Making  the  substitutions : 


T, 


0.125  X  1600000 
10.95  X   337-694 


54.1  tons. 


The  method  by  the  deflection   polygon   gives  nearly  the 
same  result,  as  will  now  be  shown. 
Comparing  the  two  equations  : 

2Rv=  T^.y, 
2niriMx  —  EI-^ .  D, 

it  is  seen  that  if  £/i  be  taken  as  the  pole  distance  in  the 
deflection  polygon,  UiriAI  must  be  the  general  expression  for 
the  load  at  any  point. 

Since  EI^  is  i,6oo,ooo  foot-tons,  CD  will  represent  it  at 
320,000  foot-tons  per  inch ;  and  that  will  be  taken  as  the  pole 
distance.     Z>  —  3,  measured  downwards,  will  be  the  load  line. 

If  the  loads  were  taken  at  10.95  x  riM,  or  lo.g^riy',  the 
lower  limit,  3,  of  the  load  line  would  not  be  on  the  diagram. 
The  loads  will  therefore  be  taken  as  ^riy'  and  a  proper  reduc- 
tion will  be  made  afterwards.     Hence,  make 


D-  I  = 

4 ''('' ' 

I   —  2  — 

4  riy' 

2   -3  = 

4  riy' 

3   -4  = 

-  4  W' ' 

4  -  5  = 

-  4  fiy ' 

5  -D  = 

-  4  riy ' 

4  X  0.285  inches. 
4  X  0.52 
4  X  0.27 

—  4  X  0.108 

—  4  X  0.528 

—  4  X  0.452 


STRESSES  IX    THE  MEMBERS. 


26  J 


The  values  of  riy'  are  taken  from  the  tabic  immediately 
above. 

By  drawing  radial  lines  from  C  to  the  points  i,  2,  3,  4,  5, 
the  polygon  Cd-^d,,d'ididr,d^d'i  is  formed  in  the  usual  manner. 

The  deflection  Dd-i  measures  2.67  inches  or  26.7  feet,  full 

size.     Since  —  =  2.74,  the  deflection  with  the  true  moment- 

4 
loads  would  be  26.7  x  2.74  =  73.16  feet,  whereas  it  should  be 
but  one-eighth  of  a  foot.    Hence,  measured  by  the  same  scale, 

the  quantities  ;/ir/J/must  be ^  of  those  taken  in  the 

^  8  X  73.16 

figure,  the  pole  distance  CD  remaining  the  same. 

Since  M  —  T^y',  there  may  be  written  for  the  point  A  : 

10.95  X  riT^y'  =  10.95  X  0.285  x  320000-^  8  x  73.16. 

As  r  =  ^,  i  =  I  and  J''  =  5.7  feet  there  results: 

^k  =  547  tons. 

The  deflection  at  other  points  might  be  used  in  the  man- 
ner already  shown  in  a  preceding  Article. 

It  is  thus  seen  that  the  constructions  are  equally  simple  in 
principle  whether  r  and  /  are  constant  or  variable. 

If  the  ends  had  not  been  fixed,  it  would  only  have  been 
necessary  to  use  the  condition 

-^^riMy^O. 

Art.  53. — Determination  of  Stresses  in  the  Members  of  an  Arched  Rib — 
Example — Fixed  Ends — Consideration  of  Details. 

It  has  been  shown,  in  the  preceding  Articles,  how  to  deter- 
mine the  horizontal  tension  7\  in  the  various  cases  which 
may  arise  ;  the  method  of  using  this  horizontal  tension  in  the 
determination  of  the  stresses  in  the  individual  members  of  an 
arched  rib  remains  to  be  shown. 

For  this  purpose  there  will  be  taken  the  rib  shown  in  PI. 


268  ARCHED  RIBS. 

X.,  Fig.  3,  having  ends  free,  /.  e.,  free  to  turn  about  the  points 
J/ and  L. 

The  curve  which  has  hitherto  been  used,  and  called  the 
'■  centre  line  "  of  the  rib,  is  the  centre  line  of  the  neutral  sur- 
face of  the  arched  rib  considered  as  a  beam  ;  consequently  the 
centres  of  gravity  of  the  various  cross  sections  of  the  rib  must 
be  found  in  this  "  centre  line,"  m  all  cases.  In  other  words,  the 
"  centre  line  "  which  has  been  used  in  the  preceding  articles 
is  the  locus  of  the  centres  of  gravity  of  the  normal  cross  sec- 
tions of  the  actual  rib. 

In  the  rib  taken  as  the  example,  PI.  X.,  Fig.  3,  the  apices 
in  the  upper  and  lower  chords  lie  in  the  concentric  circum- 
ferences of  circles  having  radii  of  78  and  72  feet,  respectively ; 
and  the  centres  of  gravity  of  the  normal  cross  sections  will  be 
supposed  to  lie  on  the  circumference  of  a  circle  having  the 
same  centre,  whose  radius  is  75  feet.  Thus  the  centre  line  is 
precisely  the  same  as  has  been  used  in  the  preceding  Articles. 
The  same  span  and  loading  will  also  be  taken. 

The  extremities  of  the  span,  or  points  M  and  L,  at  which 
the  horizontal  tension  or  force  is  applied,  imist  lie  in  the 
centre  line. 

All  the  loading  will  be  assumed  to  be  applied  at  the  apices 
of  the  upper  chord,  although  the  operations  would  be  exactly 
the  same  if  the  fixed  load  were  divided  in  any  proportion  be- 
tween the  two  chords. 

The  apices  of  the  triangles  of  the  web  system  were  located 
as  follows :  As  was  done  in  finding  Z^,  the  centre  line  was 
divided  into  ten  equal  parts.  The  upper  chord  panel  points 
are  vertically  over  those  points  of  division.  The  upper  chord 
panels  were  then  bisected,  and  radii  were  drawn  through  these 
points  of  bisection.  The  lower  chord  panel  points  were  taken 
at  the  intersections  of  those  radii  with  the  circumference  of 
the  circle  whose  radius  was  72  feet. 

There  is  only  one  point  to  be  observed  in  forming  the  chord 
panels,  i.  e.,  the  panel  points  must  be  so  located  that  the  load 
will  act  exactly  as  was  supposed  in  determining  T^. 

The  following  loads  will  then  be  assumed  to  act  through 
the  upper  chord  panel  points : 


ST/C£SS£S  IN    THE   MEMBERS.  269 

7  tons  at  the  intersection  of  i  and  3  ; 
14     "      "     "    intersections  of  4  and  5,  6  and  7,  8  and  9,  10 

and  1 1  ; 
9     "      "     "    intersection  of  12  and  13; 
4     "      "     "    intersections  of   14  and   15,  16  and   17,  18  and 

19,  20  and  21  ; 
2     "      "     "    intersection  of  22  and  24. 

As  the  ends  are  free,  L  —  \o  —  50.7  tons  in  PI.  VI.,  gives 
the  reaction  R  at  the  left  end  of  the  span,  or  at  L  in  the  ex- 
ample. Also  Z  —  I  in  PI.  VI.,  gives  the  reaction  as  30.3  tons 
at  M  in  the  example.  In  Art.  49  it  is  found  that  Tf^  —  52.5 
tons  for  this  case,  and  the  two  methods  in  Art.  51  give  the 
thermal  stresses  in  the  horizontal  tie  as  13.6  and  16.7  tons. 
The  thermal  tension  will  be  taken  at  15  tons.  The  total  ten- 
sion in  the  tie  will  then  be  52.5  +  15.00  =  67.5  tons,  as 
shown. 

It  is  a  matter  of  no  consequence  whether  the  tie  exist  or 
not.  If  it  does  not  exist,  the  abutments  at  M  and  L  must 
then  supply  the  horizontal  force  of  67.5  tons. 

Fig.  4  of  PI.  X.  is  the  complete  diagram  for  the  stresses 
with  the  load  taken  ;  it  is  drawn  to  a  scale  of  20  tons  to  the 
inch,  nearly.  The  lines  indicated  by  letters  or  figures  in  the 
diagram  are  parallel  to  the  members  of  the  rib  indicated  by 
the  same  letters  or  figures,  though  the  parallelism  is  not  ex- 
actly shown  in  the  plate  for  all  the  lines. 

With  the  following  explanations  relating  to  the  diagram, 
little  more  is  needed  : 

db  —    b'c  =  c'd'  r^d'f—  14  tons. 
on  =  ti m  =  in  k  =  k Ji  =4      " 
///■=    9      " 

q'd  =  (  l)  +  7. 
pd  =  (24)  +  2. 
PS  ^{£). 

ae  is  the  reaction   R,  and  c'd  is  the   reaction   at  M,  while 


270  ARCHED   RIBS. 

e'P'xs  the  horizontal  tension  67.5  tons.  Although  the  diagram 
appears  very  complicated,  yet,  it  is  really  composed  of  very 
simple  yf^'^-sided  figures,  as  may  easily  be  seen.  Let  the  rib 
be  divided  through  c,  7,  B,  and  T;  then  the  portion  of  the  rib 
between  that  surface  of  division  and  L  is  held  in  equilibrium  by 
the  action  of  the  stresses  in  the  members  divided  (considered 
as  forces  external  to  that  portion),  the  applied  loads,  {T),  and 
the  reaction  R.  The  resultant  of  the  loads  and  i?  is  a  verti- 
cal shear  represented  by  e'c'  in  the  diagram.  The  forces  act- 
ing upon  the  portion  of  the  rib  in  question  are  then  repre- 
sented by  the  lines  cc',  (T),  (B),  (7),  and  (c)  in  Fig.  4,  and  these 
constitute  a  simple  five-sided  figure.  The  arrow  heads  show 
the  direction  of  action  of  these  forces,  and  enable  the  kifid  of 
stress  to  be  recognized  at  a  glance. 

The  whole  diagram  is  thus  composed  of  just  such  pen- 
tagons. 

As  a  check  on  the  accuracy  of  the  construction  of  the 
diagram,  if  it  is  worked  continuously  from  L  to  J/,  the  four- 
sided  figure  involving  (7^),  (23)  and  (24)  should  exactly  close. 
It  is  far  more  conducive  to  accuracy,  however,  to  work  up 
the  diagram  from  both  ends,  and  if  the  work  has  been  accu- 
rately done,  the  diagrams  will  give  the  same  stress  in  that 
member  which  becomes  common  to  both  where  they  meet. 

The  stresses,  as  determined  by  the  original  diagram  from 
which  Fig.  4  was  constructed,  are  written  in  Fig.  3. 

If  an  arched  rib  is  subjected  to  a  load,  advancing  panel  by 
panel,  the  stresses  due  to  the  fixed  load  alone  may  first  be  de- 
termined and  then  tabulated.  The  stresses  in  all  the  mem- 
bers of  the  rib  due  to  each  panel  moving  load  may  then  be 
found  and  tabulated  also.  The  greatest  stress  of  either  kind 
in  any  member  may  then  be  determined  by  a  combination  of 
these  results  in  the  usual  manner. 

Some  of  the  stresses  found  by  diagram  should  be  checked 
by  moments  in  the  following  manner.  The  horizontal  dis- 
tances of  the  panel  points,  in  the  left  half  of  the  rib  holding- 
14  tons  each,  from  a  vertical  line  bisecting  the  span  and  pass- 
ing through  the  intersection  of  12  and  13,  are  10.9  feet,  21.7 
feet,    32    feet,    and  41.5    feet.      The   normal    distance    from 


sy/^ESSES  IN    THE   MEMBERS.  2/1 

the  intersection  of  (12)  and  (13)  to  E,  is  6.15  feet  (by  scale). 
Hence,  taking  moments  about  that  intersection, 

/  c.       507  X  50  —  14  (10.9  +  21.7+  32  +41.5)  —  67.5  X  22.1 
(^J  =  -^^ = 

—  72  tons. 


The  diagram  gave  72.5  tons,  and  the  agreement  is  suffi- 
ciently close. 

On  account  of  the  ill-defined  intersections  of  the  prolonged 
chord  sections  in  any  panel,  the  method  of  moments  for  the 
web  stresses  is  not  satisfactory  unless  one  chord  stress  in  the 
panel  is  known.  The  web  stress  can  then  be  found  by  mo- 
ments in  a  manner  to  be  presently  illustrated. 

Again,  let  (C)be  determined  by  taking  moments  about  the 
intersection  of  16  and  17.  Draw  a  vertical  line  through  that 
point.  The  horizontal  distances  of  the  three  upper  chord 
panel  points  on  the  right  of  that  line,  from  the  same,  are 
10.3  feet,  19.8  feet,  and  28.5  feet  (by  scale).  In  the  same 
manner  the  vertical  distance  of  the  point  above  7"  is  19.25 
feet.     Hence, 

30.3  X  28.5  -4(iO-3  +  19-8) -67-5  X  19-25  _ 
(^)_  __  _ 

—  90.5  tons. 

The  diagram  gave  95  tons,  and  the  agreement  is  not  close. 
This  illustrates  in  a  marked  manner  the  great  fault  of  the 
graphical  method.  In  constructing  the  original  diagram, 
shown  by  Fig.  4  of  PI.  X.,  the  rib  was  drawn  to  a  scale  of  5 
feet  per  inch,  and  the  diagram  itself  to  a  scale  of  10  tons  per 
inch,  and  although  the  greatest  care  was  taken,  yet  the  stresses 
found  for  the  right  half  of  the  rib  may,  in  some  members,  be 
wrong  to  the  extent  of  even  twenty  per  cent.  The  method 
requires  the  largest  and  most  accurate  figures  possible,  and 
the  very  nicest  instruments,  for  extended  diagrams. 

By  far  the  most  accurate,  and,  all  things  considered,  the 
most  satisfactory  method,  is  the  combination  of  moments  and 


272  ARCHED   RIBS. 

diagram,  so  freely  used  in  the  treatment  of  bowstring  trusses 
In  this  method  a  stress  in  either  chord  is  found  by  moments, 
the  other  two  stresses  (one  a  chord  and  the  other  a  web)  in 
the  same  panel  are  then  immediately  found  by  a  simple  five- 
sided  figure  or  diagram. 

The  chord  stresses  {E)  and  [G)  have  just  been  found.  In 
Fig.  I  take  bd  equal  to  67.5  tons  and  parallel  to  T,  and  make 
dc  (parallel  to  E  as  well  as  T)  equal  to  (£")  =  —  72.00  tons. 
A  section  is  supposed  to  be  taken  through  the  panel  in  which 
/",  13  and  E  are  found;  consequently  the  vertical  shear  .S  = 
50.7  —  65.00  =  —  14.3  tons.  The  remainder  of  the  diagram 
needs  no  explanation.     It  gives: 


(f)    ^ 

ah  ^  S  -14:3. tons 
dc  =(E) 

\3) 

ioL^CT) 

V;r 

(T) 

cU 

'i 

Fig.  I. 

(13)  =  —  20.3  tons  ;  (/)  =  +  17.4  tons. 

Fig.  2  is  drawn   in   precisely  the  same    manner  by  using 
\G)  =  —  90.5  tons,  which  has  already  been  determined  by 


the  moment  method.     The  section   is  taken  through  G,  17 
and  /i. 

Fig.  2  gives : 

(17)  =  —  5.5  tons ;  (/i)  =  +  23.0  tons. 

All  the  other  stresses  may  be  found  in  the  same  manner. 
The  difference  in  the  case  of  (//)  between  the  results  of  the 


ST/i£SS£S  IN   THE  MEMBERS. 


273 


two  methods  is  four  tons,  or  about  17  per  cent,  of  the  smallest 
result. 

By  the  method  of  Figs,  i  and  2  accurate  results  may  be 
obtained  by  taking  a  scale  of  even  twenty  tons  to  the  inch, 
but  a  larger  diagram  is  preferable. 

If  the  ends  of  the  rib  are  fixed,  the  shortest  method  of  find- 
ing the  stresses  is  in  no  way  different  from  that  given  in  con- 
nection with  Figs.  I  and  2,  excepting  this :  the  chord  stress 
which  is  found  by  moments  will  have  a  different  value.  If 
the  ends  are  fixed,  the  reaction  R,  at  the  left  end  of  the  span, 
will  be  L  —  \o  of  PL  V.  ;  and  the  horizontal  tension  (  T)  tvill  be 
taken  as  54.8  +  113.3  =  168. i  tons,  from  Arts.  48  and  50. 

The  line  of  action  of  T^  for  both  external  load  and  thermal 
stresses  is  H"K"  of  either  PI.  V.  or  PI.  VII.;  let  PI.  V.  be 
considered. 

The  action  of  T^  through  H"  (taken  as  acting  toward  K"), 
so  far  as  the  rib  BAD  is  concerned,  is  equivalent  to  7"^  aeting 
through  B  toward  D,  combined  zvith  a  right  hand  couple  luhose 
force  is  Tft  and  ivJiose  lever  arm  is  BH" .  Let  the  moment  ot 
this  couple  be  called  M.  This  moment  will  cause  compression 
throughout  the  upper  chord  of  the  rib  and  tension  throughout 
the  lower. 

Let  J/j  represent  the  moment  (of  the  external  forces  and 
T^)  about  any  panel  point  of  the  rib,  as  F,  Fig.  3  (the  reaction 
and  /"ft  being  taken  for  the  particular  case,  as  just  indicated). 
Then  any  chord  stress,  as  {BE),  will  be 

«i  being  the  normal  depth  of  the  rib,  as  shown.  Particular 
care  is  to  be  taken  in  regard  to  the  signs  of  J/ and  J/i,  i.  e.,  it 
is  to  be  noticed  whether  they  tend  to  produce  the  same  or  dif- 
ferent kinds  of  stress  in  BE. 

After  {BE)  is  found,  the  diagrams  are  to  be  drawn  precisely 
like  Figs,  i  and  2,  and  the  resulting  stresses,  scaled  from  the 
diagrams,  are  the  ones  desired. 

The  following,  but  longer  methods,  may  also  be  used  • 
lb 


274 


ARCHED   RIBS. 


The  first  portion  of  the  operation  is  simply  the  application 
of  the  method  by  diagram,  or  the  combination  of  moments 
and  diagram,  already  given  in  connection  with  the  case  of 
ends  free. 

All  the  individual  stresses  in  the  rib  are  to  be  found  in  this 
manner ;  tJiose  for  the  iveb  members  are  the  true  web  stresses 
desired  if  the  rib  is  of  uniform  normal  depth.  The  chord 
stresses  thus  found  are,  however,  in  all  cases,  to  be  modified. 

Let  the  normal  depth  of  the  rib  at  any  section  be  «i ;  the 

distance  BH'\  PI.  V.,  h "  ;  and  the  general  expression  for  any 

chord  stress  due  to  the  moment  M  —  T^h',  c. 

Then  will  result : 

_M_T\h^ 

Then  let  {c)  be  the  general  expression  for  any  chord  stress 
already  found  without  considering  the  moment  M  \  the  nu- 
merical value  may  be  either  positive  or  negative.  Finally, 
the  resultant  chord  stress  desired  will  be,  for  the  upper  chord, 

(c)  -  c=(c)-  ^f, 
and  for  the  lower. 

If  the  rib  is  of  uniform  normal  depth,  i.e.,  if  Wj  is  constant, 
the  web  stresses  will  not  be  affected  by  the  moment  M,  for  it 
(the  moment  M)  will  cause  uniform  chord  stresses  throughout 
the  rib. 

If  the  normal  depth,  however,  is  not  constant,  the  moment 


J/ will  cause  web  stresses  which  may  be  determined  very  ac- 
curately in  the  following  manner  ; 


STRESSES  IN    THE  MEMBERS. 


275 


Let  it  be  desired  to  determine  the  stress  in  the  web  mem- 
ber DF  of  a  portion  of  an  arched  rib,  shown  in  Fig  3,  and  let 
w  denote  that  stress.  The  stress  in  DE  is  c,  found  by  the 
method  just  given. 

In  Fig.  4,  take  {DE )  =c  and  parallel  to  DE  in  Fig. 3.  The 
lines  {GF)  and  (DF)  in  Fig.  4,  are  then  parallel  to  GF  amd 


(jos:)  -  c rojp)  -  ^ 


Fig.  4. 

DF  in  Fig.  3,  and  they  are  the  stresses  in  those  members. 
All  the  web  stresses  and  the  chord  stresses  in  one  chord  may 
be  thus  found.  This  operation  is  simply  the  method  of  Figs. 
I  and  2  applied  to  this  case. 

The  web  stresses  may  be  found  by  using  moments,  only,  in 
the  following  manner  : 

Take  ^  as  any  convenient  point  in  GF.  Let/]  represent 
AB,  and  4,  AC;  these  lines  are  normal  to  DE  and  DF  re- 
spectively. 

Let  M  be  still  considered  right-handed  and  positive,  and 
let  it  first  be  assumed  that  c  is  compression.  Moments  about 
A  give  : 

_  _  -r/i  +  T„h' 

If  c  is  tension,  or  belongs  to  a  panel  in  the  lower  chord, 
then  {w)  will  be  the  stress  in  a  member  like  DG.  There  will 
then  result  : 

—  cli  +  TJt" 

w  —  — — - —  . 

4 

In  either  of  these  formulae,  (w)  will  represent  tension  or 
compression  according  as  the  result  is  positive  or  negative. 

Let  {w)  represent  any  web  stress  already  found  by  neglect- 
ing M,  then  will  any  resultant  web  stress  desired  be : 

{w)  +  w\ 

the  signs  of  both  these  quantities  being  implicit. 


2^6  ARCHED  RIBS. 

As  the  lever  arms  /, ,  A ,  and  Wj  are  scaled  from  the  draw- 
ing, the  rib  should  be  laid  down  as  accurately  and  to  as  large 
scale  as  possible. 

In  important  cases  these  different  methods  should  be  used 
as  checks. 

A  very  common  system  of  bracing  for  arched  ribs,  although 
a  very  unsatisfactory  one,  is  that  shown  in  Fig.  5.     The  web 


Fig.  5. 

members,  bB,  cC,  dD,  etc.,  are  normal  to  the  centre  line  of 
the  rib,  and  are  designed  for  tension  only.  The  other  web 
members  are  for  compression  only 

Let  the  ends  be  supposed  free,  and  take  AOP  for  the  true 
equilibrium  polygon  for  a  given  load. 

For  any  given  loading  the  stresses  in  the  different  members 
are  indeterminate,  unless  about  half  of  the  compression  web 
members  are  neglected. 

With  the  assumed  position  of  the  equilibrium  polygon  AOP, 
for  instance,  it  is  seen  that  compression  will  increase  in  the 
upper  chord  from  b  io  d  (nearly) ;  from  that  point  to  g 
(nearly)  it  will  decrease.  The  compression  in  the  lower  chord 
will  increase  from  G  to  L  (nearly)  and  then  decrease  from 
that  point  to  N.  The  points  of  greatest  chord  stresses  of 
either  kind  are  those  at  which  the  polygon  and  centre  line  of 
rib  are  parallel. 

From  these  considerations  it  results  that  the  web  members 
bC,  cD,  De,  Ef,  Fg,  Gh,  Hk,  Kl,  IM,  and  mN  may  be  omitted  ; 
they  must  be  omitted,  in  fact,  if  the  stresses  are  to  be  deter- 
minate. Having  made  these  omissions,  the  stresses  are  to 
be  found  by  the  methods  already  given,  as  those  methods. 
.  are  perfectly  general. 


STRESSES  LV    THE   MEMBERS. 


277 


Precisely  the  same  observations  apply  to  the  case  of  fixed 
ends,  or  to  that  in  which  the  normal  members  are  in  com- 
pression and  the  others  in  tension. 

It  is  by  no  means  certain  that  the  stresses  thus  found  will 
really  exist  in  the  rib,  but  the  assumption  is  the  best  that 
can  be  made.  This  system  of  bracing  is,  at  best,  very  un- 
satisfactory. 

The  free  ends  of  an  arched  rib  are  sometimes  arranged,  in 
regard  to  support,  as  shown  in  Fig.  5,  PI.  X.  There  are  two 
ties  or  sets  of  ties,  T'  and  T'\  instead  of  one,  7),.  A  and  B 
are  the  points  at  which  these  ties  take  hold  of  the  rib.  E  is 
the  intersection  of  AB  and  the  centre  line,  EF,  of  the  rib. 
The  span  to  be  used  in  finding  T,,  for  either  external  loads  or 
thermal  variation  is  the  horizontal  distance  between  E  and  the 
corresponding  point  at  the  other  end  of  the  rib.  T\  T^,  and 
T"  are  parallel  to  each  other,  and  ac  is  normal  to  the  three. 

Now  if  T'  and  T"  are  determinate,  there  may  be  written  : 

T'  --T  ' 

^     -  ac  ^'^• 

In  such  a  case  the  systems  of  triangulation  in  the  rib  may 
be  separated,  T'  will  belong  to  one  and  T"  to  the  other. 
The  stresses  may  then  be  found  in  each  system  separately, 
and  the  results  combined  for  the  resultant  stresses  of  the  rib. 
If  the  web  members  may  be  counterbraced,  the  resultant 
stresses  are  thus  determinate.  It  is  not  certain,  however, 
that  the  tensions  T'  and  T"  will  have  the  values  given  above. 

For  the  determination  of  the  stresses  in  the  rib,  however, 
it  is  not  necessary  to  resolve  Tj^  into  T'  and  7"",  except  for 
the  panel  ABDC. 

In  the  case  of  a  design,  if  the  dimensions  required  by  the 
calculations  of  this  Article  give  a  value  to  the  moment  of 
inertia  /  very  different  from  that  assumed  in  the  determina- 
tion of  7"/^,  for  either  thermal  variations  or  external  load,  it 
will  be  necessary  to  make  an  entirely  new  set  of  calculations- 


2y8 


ARCHED   RIBS. 


with  another  value  of  /.  This  must  be  done  until  the  agree- 
ment between  the  assumed  and  required  values  of  /  is  suf- 
ficiently close. 


Art.  54. — Arched  Rib  Free  at  Ends  and  Jointed  at  the  Crown, 

Suppose  the  rib  to  be  represented  in  the  figure. 

Since  there  is  a  joint  at  A,  the  bending  moments  must  be 
zero  at  that  point ;  consequently  the  equilibrium  polygon  for 
any  load  must  pass  through  that  point.     This  fact  furnishes  a 


Fig.  6. 


very  simple  method  of  determining  7^.  Denote  by  "2 Px  the 
moment  of  all  the  external  forces  about  the  joint  A  ;  then, 
since  J/ must  be  equal  to  zero, 


^Px-  T^{AC) 


T,= 


^Px 

AC 


In  this  rib  variations  of  temperature  produce  no  variations 
of  stress  in  BD,  except  that  due  to  the  slight  change  of  A  C, 
as  shown  by  the  formula  above.  This,  however,  is  a  very 
small  quantity,  and  would  ordinarily  be  neglected.  If  neces- 
sary, it  would  be  allowed  for  by  taking  the  value  oi  ACat  the 
lowest  temperature  to  which  the  rib  would  be  subjected. 

Other  arched  ribs  are  seldom  constructed,  but  they  are  to 
be  treated  by  the  same  general  methods,  precisely,  as  those 
used  in  the  preceding  cases. 


CHAPTER    IX. 


SUSPENSION   BRIDGES. 


Art.  55. — Curve  of  Cable  for  Uniform  Load  per  Unit  of  Span — Suspen« 
sion  Rods  Vertical — Heights  of  Towers,  Equal  or  Unequal — Gen- 
eralization. 


Fig.  I. 


In  the  figure,  let  EH' C  represent  the  cable  of  a  suspension 
bridge  carrying  a  load  extending  over  the  whole  span.  In 
the  ordinary  experience  of  an  engineer,  the  load  carried  by  a 
suspension  bridge  cable  is  nearly  uniform  in  intensity  in 
reference  to  a  horizontal  line ;  so  nearly  uniform  per  foot  of 
span,  in  fact,  that  it  is  assumed  to  be  exactly  so,  and  such  an 
assumption  will  be  made  in  the  present  instance. 

The  use  of  the  stiffening  truss,  to  be  presently  noticed, 
makes  this  assumption  essentially  true. 

Let  {ED  +  BC)  =  I  =  span  ;  BH'  =  h^  ;  DH'  =  Jh  \ 
w  =  load  for  horizontal    foot,  and   let  x  be   measured  hori- 

279 


28o  suspensiojv  bridges. 

zontally  from  H',  the  lowest  point  of  the  cable.     The  height 
of  the  highest  tower  is,  of  course,  h^,  and  that  of  the  other  //j. 

The  ordinate  of  any  point  P  is  x,  the  load  on  H'M  is,  con- 
sequently, W=  wx.  Draw  PK  tangent  to  the  curve  at  P, 
then  by  the  first  principles  of  statics,  it  is  known  that  the 
direction  of  the  cable  tensions  at  /"and  H'  and  the  direction 
of  W  must  intersect  in  one  point  N.  Since,  however,  %v  is 
uniform  along  x,  the  resultant  direction  of  W  passes  through 
N,  half  way  between  H'  and  M.  Hence  FH'  =  H'K;  or, 
since  FK  \s,  the  snhidingent,  the  abscissa,  FH',  of  the  curve  is 
equal  to  half  the  subtangent,  consequently  the  curve  is  the  or- 
dinary parabola. 

Again,  it  is  known  that  the  horizontal  component  of  the' 
tension  of  a  cable  will  be  a  constant  quantity  if  the  loading 
(as  in  the  present  case)  be  wholly  vertical ;  let  that  compo- 
nent be  denoted  by  H. 

Let  GNP  be  taken  for  the  triangle  (right  angled)  of  forces 
at  P,  in  which  NP  represents  the  cable  tension  at  P,  GN  the 
load,  W^  wx  and  C/'the  constant  horizontal  component  H. 

Then  let  AP  be  drawn  normal  to  the  curve  at  P\  the 
triangles  AFP  and  GNPv^WX  be  similar.  There  can  now  be 
at  once  written  the  relation  : 


AF 

FP        x         I 

GP~ 

GN      wx       w ' 

H    .' 

TT 

.     AF  —  —  =  constant 

w 

but 

GP  =  H    .'.     747^=—  =  constant     .     .     (i). 

w 

Now  AF  is  the  subnormal  of  the  curve  of  the  cable,  and 
since  it  is  constant,  the  curve  is  the  ordinary  parabola. 

The  preceding  results  may  be  generalized  in  a  very  simple 
and  easy  manner. 

If  any  two  points,  as  P  and  Q,  be  considered  fixed,  and  if 
the  portion  PQ  of  the  cable  carry  the  same  intensity  of  load 
w  as  before,  there  will  at  once  result  the  general  case  of  a 
flexible  cable  carrying  a  load  whose  intensity,  along  a  straight 
line,  and  direction  are  uniform.     There  may  then  be  stated 


PARAMETER    OF  CURVE.  28 1 

the  general  principle  :  If  a  perfectly  flexible  cable  carry  a  load 
2iniforin  in  directio?i  and  intensity  in  reference  to  a  straight 
lifie,  the  cable  will  assume  the  form  of  an  ordinary  parabola 
whose  axis  zvill  be  parallel  to  the  direction  of  the  loading. 

This  principle  finds  its  application  in  the  case  of  a  suspen- 
sion bridge  with  inclined,  but  parallel,  suspension  rods. 


Art.  56. — Parameter  of  Curve — Distance  of  Lowest  Point  of  Cable  from 
either  Extremity  of  Span — Inclination  of  Cable  at  any  Point. 

Attending  to  the  figure  and  notation  of  the  previous  Ar- 
ticle, the  equation  of  the  curve,  the  origin  of  co-ordinates 
being  taken  at  //',  is: 

x^  —  2py ; 

in  which  2/  is  the  parameter. 

Let  BC  =  Xi  and  ED  —  Xo,  then  there  may  be  written  Xi^  = 
2phi,  x-i  =  2ph2  and  ix^x.^  =  ^p\/h^h<^. 

Hence 

{x^^x^f=l^=2p{Vh',+Vh:,f=2p{h,  +  2VT,h^  +  h.;)  .   .  .  (I). 

.'.   P  = ^ = =     ==:= .    .    (2). 

2  ( Vhi  +  V/h  f         2(//i  -I-  2  V/h  fh  +  /^2) 
If  the  towers  are  of  the  same  height,  then  //j  —  li^  =  h  and 

^=£ (3). 

Now  Xx  is  the  horizontal  distance  from  the  lowest  point  of 
the  cable  to  that  end  of  the  span  at  which  //^  is  found,  /.  e., 
BC  in  the  figure,  while  x^  is  the  other  segment  of  the  span, 
and  by  the  equations  immediately  preceding : 

^1=     /T-^    /7- (4)- 


232  SUSPENSION  BRIDGES. 

/ 

If  //i  =  //2 ;  -n  =  jTa  =  -     •     •     .     .     (6). 

Referring  to  the  figure,  since  KH'  =  H'F  =  y,  if  i  is  the 
incHnation  to  a  horizontal  Hne,  of  the  curve  at  any  point  Py. 
then  X  tan  i  =  2y  ; 


21/        .    .  /     4y 

hence,  tan  i  —  -^  .'.  sec  i  —  \/  i  +  -~V 


At  the  summits  of  the  towers 


t/' 


(7). 


tan  i\  =  ~—^  and  tan  i^  =  — -    ....     (8). 
Xi  x^ 

\i  hi  =  /^2,  ta)i  ii  =  tan  1^=  .     .     .     (9). 


Art.  57. — Resultant  Tension  at  any  Point  of  the  Cable. 

In  the  first  Article  of  this  Chapter  there  was  recognized  the 
general  principle  that  if  the  loading  on  a  cable  is  uniform  in 
direction,  the  component  of  cable  tension  normal  to  that 
direction  will  be  constant  at  all  points  of  the  cable.  In  the 
present  case  the  resultant  tension  at  the  lowest  point  of  the 
cable  will  be  this  constant  component  H. 

From  Eq.  i,  Art.  55,  //=  zuAF.  But  AF  is  the  subnormal 
of  the  curve,  and,  from  Analytical  Geometry,  it  is  known  to 
be  equal  to  one  half  of  the  parameter,  or  equal  to/  (using  the 
same  notation  as  before). 

Hence,  after  taking  the  value  of  p  from  the  previous 
Article: 

H=wp  = -.= 7^—  = , •   (0- 

2  ( V //i  +  V //2  f        2  (h■^  +  2  V //i  /t^  +  A,) 


LENGTH  OF  CURVE.  283 

Let  R  denote  the  resultant  tension  at  any  point,  then  by  the 
triangle  of  forces  GNP,  in  the  figure : 


R=  H  seci=  HA/  1  +  ^  .    .     .     .     (2). 

Eq.  (2)  gives  the  tension  at  any  point.     At  the  summits  of 
the  towers  there  arc  found  : 


^i  =  //j/i  +  ^i- (3). 


/' 


R,  =  Hi/i^^ (4). 


If  hi  =  //j,  consequently  Xx  =  X2  =  — ,  then  : 


H='il,R,^R,  =  H^,^'-^    .    .     .     (5). 


S& 


Art.  58. — Length  of  Curve  between  Vertex  and  any  Point  whose  Co- 
ordinates are  x  and  y,  or  at  which  the  Inclination  to  a  Horizontal 
Line  is  /. 

The  usual  expression  for  the  length  of  a  part  of  one  branch 
of  a  parabola,  beginning  at  the  vertex,  as  determined  by  the 
integral  calculus,  may  easily  be  put  in  the  following  form, 
denoting  by  c  the  length  in  question  : 


x" 
c  = 
A}' 


:,l?'(/-^-Hyp.,Og.(^/y,.^)|(.). 


Or,  using  the  values  for  tan  i,  sec  i,  and  /,  determined  in 
the  preceding  Articles : 

c  =  -<-  \tan  i  sec  i  4-  hyp.  log.  {tan  i  -\-  sec  i)\     .     .     .      (2). 


284 


SUSPENSION  BRIDGES. 


The  total  length  of  the  cable  will,  of  course,  be  found  by 
putting  x\  and  hy  for  x  and  y  in  Eq.  (i),  or  Zj  for  i  in  Eq.  (2) ; 
then  X2  and  fi^  for  x  and  j/,  or  ^2  for  /,  and  adding  the  results. 

Denoting  those  results  by  fj  and  c^  the  total  length  will 
then  be : 

Ci  +  c>. 

An  approximate  formula  sometimes  used  is  determined  as 
follows.  In  the  figure  of  the  first  Article  of  this  chapter  con- 
sider H'Pto  be  the  arc  of  a  circle,  and  let  x  and  y  be  taken 
as  heretofore  ;  also  let  R  be  the  radius  of  the  circle.  The 
ordinary  expression  for  the  length  of  a  circular  arc,  in  the 
integral  calculus,  is  : 

(nearly), 


if  X  is  small  compared  with  R.  Again,  performing  the  divi- 
sion indicated  and  omitting  all  terms  in  the  quotient  after 
the  second,  there  will  result  : 


/ 


^-(i+2^'^-^)   =^0  +6^,^     •     •    (3^- 


If  j/^  be  omitted  in  the  expression  x^  =  2Ry  —  jj^,  and  the 
resulting  value  of  R  be  inserted  in  Eq.  (3),  there  will  at 
once  be  found : 

2y^\ 


As  before,  to  find  the  total  length  of  the  curve,  x^  and  /i^, 
and  X2  and  /12  must  be  inserted  in  succession  in  Eq.  (4),  and 
the  results  added. 

If  the  heights  of  the  towers  are  equal  to  each  other  and  to 
k,  the  total  length  will  be 

<..f)   ...  ...(s). 


DEFLECTION  OF  CABLE.  28; 

The  expressions  (4)  and  (5)  are  evidently  not  close  approxi- 
Tnations  except  for  very  flat  curves,  in  which  case  the  nature 
of  the  curve  is  a  matter  of  indifference. 


Art.  59. — Deflection  of  Cable  for  Cheinge  in  Length,  the  Span  Remaining 

the  Same. 

The  approximate  formula  (4)  of  the  preceding  Article  is 
usually  used  in  determining  the  deflection. 
The  total  length  of  the  cable  is  : 

Ci+  C2  =  x\  4-  x^  +-[—  +  -r )  , 
Differentiating : 

</(.,  +  .,)  =  f  ('A +MrfA*    .    .    .    (I). 


a,,^iAki±^ (2). 

X]_         Xx 


The  variation  in  the  length  of  the  cable,  whether  arising 
from  variation  in  temperature  or  any  other  cause,  is  to  be 
put  for  d{c-^  +  c-^  in  Eqs.  (i)  and  (2),  then  dh  will  be  the  cor- 
responding deflection  of  the  lowest  point  of  the  cable. 

If  the  towers  are  of  the  same  height,  and,  coHsequently, 


2dc^  =  ^—      ^j-.dh (3). 

^^=IW//'^"^ (4). 


*  Since  h^  —  h-i  =  constant,  dhx  =  dh.^  —  dh. 


286  SUSPENSION  BRIDGES. 

It  is  assumed  in  Eqs.  (i)  and  (2)  that  the  lowest  point  of 
the  cable  remains  at  the  same  horizontal  distance  from  the 
ends  of  the  span,  though  such  is  not  really  the  case. 

The  true  deflection  can  only  be  found  by  trial  by  the  use 
of  Eq.  (i)  of  the  previous  Article. 

Let  (ci  4-  ^"a)  be  the  known  length  of  the  cable  before  varia- 
tion in  its  length  takes  place  ;  then  let  //j,  h^,  -I'l  and  x^  be  the 
original  heights  of  towers  and  segments  of  span,  also  known. 
Let  ji  and  y-^  be  the  heights  of  towers  above  the  lowest  point 
of  the  cable  after  the  variation  in  its  length  has  taken  place ; 
and  let  it  be  assumed,  as  before,  that  x^  and  x^  remain  the 
same  whatever  the  deflection. 

Let  V  be  the  variation  in  length  of  the  cable. 

Then,  since  z'  =  —  [c^  +  c^  +  {c^  -\-  c^  -\-  v): 

+  |/i  +  ^)  I  -  (^1  +  ^.)  .  .  •  (5). 

But  there  is  also  the  equation  of  condition  : 

}'i  —  yi~  ^h  ~  fh  —  constant    .     .     .     (6). 

The  value  of  y^  or  y^  may  be  taken  from  Eq.  (6)  and  put  in 
Eq.  (5),  there  will  then  be  but  one  unknown  quantity  in  the- 
right  member  of  that  equation,  and  its  value  must  be  found 
by  trial.  The  first  value  of  y^  or  y.,  taken  may  be  /i^  or  ^2 
increased  or  decreased,  as  the  case  may  be,  by  d/i  given  by 
Eq.  (2). 

The  deflection  sought  is,  of  course : 

yi  — /ix  =  y«  - /ti 

If  the  new  heights,  y^  and  y^  are  given,  the  variation  of 
length,  V,  will  be  at  once  given  by  Eq.  (5). 


Si'SPE.VSJOX   CANTI-LE  VERS. 


287 


If  heights  of  towers  are  the  same,  Eq.  (6)  will  not  be 
needed  ;  for  making  x-^  —  x^^  -,  ci=  c^,  and_>'i  —  y^_  =  h,  there 
results : 


2/2    (  4^ 


"^-xWi   { 


^|ATLf^.h3,p.,o,(^./77^)[ 


-2^1 *  (7)- 


In  Eq.  (7)  /z  is  then  to  be  found  by  trial,  as  before,  if  v  is 
given  ;  or  if  Ji  is  given,  v  at  once  results. 

The  deflection  of  the  middle  point  of  the  truss  will  be  : 

//  -  //i. 

It  is  to  be  noticed  that  in  Eqs.  (5)  and  (7)  all  the  quantities, 
ji,  j'2,  and  //,  increase  in  the  same  direction  with  v.  This  ma- 
terially simplifies  the  approximation  by  trial. 

The  determination  of  v  in  Eq.  (5)  might  be  made  without 
assuming  ^1  and  x^  to  remain  constant,  for  there  are  two  other 
equations  of  condition : 

.f  1  +  X2  ^=  /, 
and 

These,  with  Eqs.  (5)  and  (6)  would  be  sufficient  in  order  to 
find  the  four  unknown  quantities,  j'^,  j.^,  Xi  and  ^2 . 

Such  a  degree  of  extreme  accuracy,  however,  is  unneces- 
sary. 


Art.  60. — Suspension  Canti-Levers. 

In  the  figure,  ABD  represents  a  suspension  canti-lever.  The 
cable  BC  goes  over  either  to  another  span  or  to  an  anchorage, 
while  A  is  the  end  of  the  canti-lever.  The  cable  AB  is  in 
precisely  the  same  condition  as  the  half  of  a  cable  belonging 


288 


SUSPENSION  BRIDGES. 


to  a  span  equal  to    2AD;  consequently  its  tension  R  at  any 
point  and  its  inclination  at  the  same  point  are   to  be  found 

by  the  formulas  already 
given.  In  fact,  all  the 
circumstances  are  pre- 
cisely the  same  except 
this,  the  platform  is 
subjected  to  a  thrust, 
uniform  throughout  its 
whole  length,  and  equal 
to    the     constant    hori- 


FlG.    I. 


zontal  component,  //,  of  the  tension  R. 

Art.  61. — Suspension  Bridge  with  Inclined  Suspension  Rods — Inclination 
of  Cable  to  a  Horizontal  Line — Cable  Tension — Direct  Stress  on  Plat- 
form— Length  of  Cable. 

In  this  case  the  suspension  rods,  or  suspenders,  are  all  sup- 
posed to  be  equally  inclined  to  a  vertical  or  horizontal  line, 
and,  consequently,  are  parallel  to  each  other  ;  they  are  also 
supposed  to  take  hold  of  the  platform  or  stiffening  truss  at 
points  equidistant  from  each  other.  These  conditions  cause 
the  cable  to  be  subjected,  in  each  of  its  parts,  to  the  action 
of  parallel  loading  of  uniform  intensity  in  reference  to  the 
span.  As  was  shown  in  the  first  Article  of  this  Chapter,  the 
curve  of  the  cable  will  be  composed  of  common  parabolas 
having  axes  parallel  to  the  suspension  rods. 

In  the  figure,  A  is  the  lowest  point  of  the  cable,  while  BD 
and  FG  represent  suspension  rods  on  either  side  of  A.     The 


angle  a  is  the  common  inclination  of  all  suspenders  to  a  ver- 


IXCLIXED   SUSPENSION  RODS.  289 

tical  line,  it  also  represents  the  inclination  of  the  axis  of 
either  of  the  parabolas  AC  ox  AE  to  the  same  line. 

The  vertex  of  the  parabola  y^C  is  on  the  left  of  A,  and  the 
vertex  of  AE  is  on  the  right  of  the  same  point. 

Let  X  be  horizontal  in  direction  and  measured  from  J,  and 
let  the  length  of  any  suspender,  as  BD  be  denoted  by  j,  but 
let  CP  be  designated  by  )\.  Either  parabola,  as  A  6,  will  then 
be  referred  to  oblique  co-ordinates  in  the  usual  manner. 

If  OB  be  drawn    tangent  to   the  curve  at  B,  AO  v^\SS.  be 

X 

equal  to  OD,  or  — .    If  i  represents  the  inclination  of  the  curve 

at  any  point,  as  B,  to  a  horizontal  line,  the  triangle  OBD  will 
give: 

BD  _2y  _       sin  i  sin  i 


OD        X       cos  (a  +  z)       cos  a  cos  i  —  sin  a  sin  V 

2y 
— ^^—cos  a 

.'.     tan  i= .      .     .     (i). 

2y      .  ^  ^ 

I  +  ~^^  sin  a 

X 


In  the  usual  manner,  sec  i  =  Vi  +  tan^  i,  or, 


a/  I  +  ^^  sin  a 


■    4/ 


sec  I  =  '      .     .     (2). 

2y      .  ^  ^ 

I  H ^^  stfi  a 

X 


At  the  point  (7,  if  CM  =  /i^,  AP=  x^,  and  AM=  a,  in  Eqs. 
(l)  and  (2),  hi  sec  a  is  to  be  put  ior  y,  and  {a  —  h^tan  a)  for;jr. 

Exactly  similar  equations  apply  to  the  other  portion  of 
the  span. 

For  the  point  A,  Eqs.  (i)  and  (2)  apparently  become  inde- 
terminate, but  only  apparently,  for  the  relation. 


y  hi  sec  a  . 

l?"'\a-  hi  tan  af  '       *     '     *     •     U> 


19 


290  SUSPENSIOA'  BRIDGES. 

gives, 

2y  _        2^1  sec  a 


X       («  —  h]^  t<^n  of    ' 
and  when  x  =  o,  consequently,    -  becomes  zero,  making  tan  i 

X 

equal  to  zero  also. 

If  OBD  be  taken  as  a  triangle  of  forces,  OB  will  be  the 
cable  tension  R  dX  B  \  while  OD  will  be  the  horizontal  com- 
ponent H,  and  BD  will  represent  wx  ser  a.  w  is  the  total 
load  per  unit  of  span  on  AD. 

From  the  triangle  in  question, 

H        _  cos  {a  -'r  i)  _    X  ^     tJ  —   "^^^'^  ^^^  ^ 


wx  sec  oc  sin  t  2y  2y 

rr       w{a  —  lu  tan  a^-  ,  . 

•••  ^=^ — - <4). 

As  was  to  be  expected,  Eq.  (4)  shows  //  to  be  a  constant 
quantity,  but  it  is  not  a  rectangular  component  in  this  case. 

The  same  triangle  gives  for  the  resultant  tension  at  any 
point : 


R=  sjiywx  sec  of  -\-H^  +  2wx  H  tan  a.    .     .     (5). 

For  the  point  C,  x  becomes  (a  —  h^  tan  ex). 
If  /  is  the  span,  these  equations  apply  to  the  other  portion 
of  it,  by  taking  //^  for  //j,  and  (/  —  a)  for  a. 

If  the  towers  are  of  equal  heights,  //j  becomes  equal  to  h^, 

and  a  =  I  —  a  =  —. 

2 

'Let  p  be  the  horizontal  distance  between  any  two  suspend- 
ers, then  the  tension,  /,  in  the  suspender  will  be : 

/  =  wp  sec  a (6). 

The  direct  stress  in  the  platform  is  caused  by  the  horizon- 
tal component  of  the  tension  in  the  suspension  rods.  This 
stress  may  exist  as  tension  in  the  platform,  in  which  case  it 


INCLINED   SUSPENSION  RODS.  29 1 

^vill  exert  no  action  on  the  towers.  Remembering  that  all 
the  suspension  rods  must,  at  any  instant,  be  subjected  to  a 
uniform  stress,  it  is  evident  that  the  direct  tension  in  the  plat- 
form will  have  its  greatest  value  at  the  centre,  and  will  be 
equal  to 

7it  sin  a  —  mvp  tan  a  ; 

in  which  ;/  is  the  number  of  suspension  rods  in  each  half  of 
the  span,  supposing  towers  to  be  of  equal  heights.  If  «'  be 
the  number  of  suspenders  between  the  end  of  the  span  and 
any  point,  the  tension  in  the  platform  at  that  point  will  be 

fit  sin  a  =  nwp  tan  a. 

If  the  towers  are  of  unequal  heights,  there  will  be  a  greater 
number  of  suspenders  on  one  side  of  the  lowest  point  of  the 
cable  than  on  the  other.  Let  n^  be  the  number  in  that  por- 
tion of  the  span  adjacent  to  the  highest  tower,  and  n^  the 
number  in  the  other  portion  ;  n^  will  be  greater  than  n.^.  In 
this  case,  then,  the  platform  at  the  foot  of  the  highest  tower 
will  sustain  a  thrust  given  by  the  expression 

(Wi  —  n-^  t  sin  a  =  {n^  —  n^)  ivp  tan  a. 

If  the  platform  is  to  sustain  a  direct  thrust  only,  at  the  feet 
of  the  two  towers  it  will  have  to  sustain  thrusts  given  by  the 
-expressions 

Tilt  sin  a  =  n^ivp  tan  a 
n^t  sin  a  =  n-fcvp  tan  a. 

If  tt  represents  the  number  of  suspension  rods  between  the 
centre  and  any  point,  the  thrust  at  that  point  will  be 

n't  sin  a  =  liwp  tan  a. 

In  the  case  of  a  suspension  cantilever,  in  addition  to  the 
thrust  given  above  there  will  be  one  denoted  by  H,  uniform 
throughout  its  length.  Other  calculations  for  a  suspension 
cantilever  are  precisely  the  same  as  those  already  given. 


292 


SUSPENSIOJ^  BRIDGES. 


The  length  of  the  cable  from  the  lowest  point  to  any  other 
point  at  vvhicli  the  inclination  to  a  horizontal  is  i,  is  readily 
found  by  means  of  the  formula  used  for  the  cable  with  verti- 
cal rods.  In  the  present  case  the  inclination  of  the  cable  at 
any  point  to  a  line  perpendicular  to  the  axis  of  the  parabola 
is  {i^-  a) ;  consequently  there  is  simply  to  be  found  the  length 
of  the  parabolic  arc  between  the  points  at  which  the  incli- 
nations to  the  axis  are  (90  —  {i+a) )  and  (90  —  a). 

The  formula  mentioned  then  gives 


c 

2\ 


—  {tan  {i+oi)  sec  {i-{-a)  —  tan  a  sec  a  -f 

,         ,        tan  (i  + «)  +  sec  (i  +a)\  ,  , 

hyp.  log. ^^ ' ^^ )     .     .     (7). 

^^       ^  tan  a  +  sec  a  J  ^' ' 

It  is  known  from  analytical  geometry  that  p  takes  the  fol- 
lowing form  in  terms  of  the  oblique  co-ordinates  used  in  this 


case: 


"  ~       2y       ~  2hx 

Eq.  (7)  is,  of  course,  to  be  applied  to  both  branches  of  the 
curve  to  obtain  the  total  length. 

From  what  was  said  in  the  demonstration  of  the  approxi- 
mate formula,  it  may  be  seen  that  it  can  be  applied  to  the 
present  case  by  changing  x  \.o  {x  -^  y  sin  a)  and,  y  to  y  cos  a. 
The  formula  then  becomes  : 

2f  co^  a 

c  =  X  +  y  sin  a  -{ -. , -. r      .      .      (o). 

3  (;r  +  J  sm  a) 


Art.  62. — Suspension  Rods;   Lengths,  and  Stresses. 

In  the  following  calculations  it  is  virtually  assumed  that  the 
cable  lies  in  a  vertical  plane,  and  that  the  suspension  rods  are 
vertical.  This,  however,  does  not  affect  the  generality  of  the 
results  obtained,  for  in  all  cases  the  suspension  rods  are  sup- 
posed parallel  to  each  other,  and  the  lengths  found  by  the 


SUSPENSION  RODS.  293 

formulae  of  this  Article  are  to  be  taken  as  the  vertical  pro- 
jections of  the  true  or  actual  lengths.  The  true  lengths  are 
therefore  to  be  found  by  multiplying  the  values  of  /io,  //i,  //^j, 
etc.,  by  the  secant  of  the  common  inclination  to  a  vertical 
line,  of  the  suspension  rods. 

Since  a  flat  parabola  nearly  coincides  with  a  circle,  the 
camber  may  be  supposed  to  be  formed  by  a  parabolic  arc. 
Let  the  co-ordinate  x  be  measured  from  A  toward  B,  in  Fig. 
14,  PL  XII.,  and  y  perpendicular  to  it ;  also  let  AB  =  x\  = 
half  span.  Then  since  the  curve  of  the  cable  is  supposed  to 
be  a  parabola  in  a  vertical  plane  : 

In  the  same  manner  for  the  camber: 

Then  the  total  length  of  any  suspender  is: 

h  =  y  +  y  +  c. 

When  the  suspenders  are  separated  by  a  constant  distance, 
d,  simpler  formulae  may  be  found. 

Each  suspender  is  composed  of  the  sum  of  two  variable 
lengths  (y  and  j")  and  a  constant  length,  <:.  Now  if  {j^  +  6) 
be  written  for  y^ : 

J=  (j'l  +  <^)— 2, 

will  evidently  be  the  sum  of  the  two  variable  lengths  referred 
to.  Hence,  if  //q,  /h,  Jh,  7^3  *  *  *  //„  represent  the  lengths 
of  the  suspenders  as  shown  in  the  figure : 

h  =  ^, 

h  =  C  +      -i  (Ji  +  «^)» 
-^1 


294  *  SUSPENSIOA'^  BRIDGES. 

h  =  c  +  —2-  ( Ji  +  6), 

hz  =  c  +  ^  (ji  +  S), 
{n-ifd\ 


Having  computed  the  lengths  of  the  suspenders  for  one 
half  the  span,  the  results  may  be  used  for  the  other  half  if  the 
piers  are  of  the  same  height  ;  otherwise  the  lengths  must  be 
computed  separately. 

The  stress  in  any  suspension  rod  is  the  vertical  load  which 
it  carries,  multiplied  by  the  secant  of  its  inclination  to  a 
vertical  line. 

Art.  63. — Pressiire  on  the  Tower — Stability  of  the  Latter — Anchorage. 

Let  P^  —  vertical  component  of  pressure  on  tower  head. 

''     A  =  horizontal 

"     7?  =  resultant 

"      Tj,  and  7^',  Fig.  13,  PI.  XII.,  =  tensions  of  the  cable  on 
different  sides  of  the  pier  head. 

a,  a',  and  d  represent  inclinations  to  the  vertical  as  shown. 
When  friction  on  the  saddle  is  considered : 

Pj,  =  Tp  cos  a  +  Tp'cos  a' ;  P^=  Tp  sin  a  —  Tpsin  a'; 


R=VPv'  +  Pk';     cos  6  =  ^. 

When  friction  on  the  saddle  is  not  considered,  Tp  =  Tp  ; 
.'.  P^=  Tp  {cos  a  +  cos  a') ;     Pj^=  Tp  {sin  a  —  sin  a')     .    (i). 

j^=VpJTW;  cose  =  ^. 


PRESSURE    ON   THE    TOWER.  295 

In  the  same  case  \{  a  =  a  ; 

P^^2W,  /\  =  o,  R^2W,  8  =  0.  {W=^  weight  of  load 
and  structure.) 

There  are  two  cases  in  which  the  resultant  pressure  on  the 
tower,  caused  by  the  tension  in  the  cables,  may  be  vertical  in 
direction.  Both,  however,  are  founded  on  the  single  condi- 
tion that  the  horizontal  components  of  the  cable  tension,  on 
each  side  of  the  tower  head  are  equal  to  each  other. 

This  condition  will  exist  i{  a  =  a'  in  Eq.  ( i ),  making  /\  =  o ; 
or  if  the  saddle  be  supported  on  rollers  and  roller  friction  be 
omitted.  In  the  latter  case  P^  =  o  because  Tp  sin  a  = 
Tp  sin  a ,  and  not  because  a  necessarily  equals  a'. 

In  discussing  the  stability  of  position  of  masonry  towers, 
let  the  distance  of  the  centre  of  pressure  from  the  centre  of 
figure  of  the  section  of  the  pier  be  denoted  by  q' .  If  this 
latter  does  not  exceed  g  (the  limit  of  safety  for  q'),  which 
may  be  ascertained  by  determining  the  line  of  resistance  for 
the  pier,  stability  of  position  will  be  secured. 

It  is  supposed,  of  course,  that  6  has  some  value  greater  than 
zero  ;   otherwise  q'^=  O. 

The  stability  of  friction  for  masonry  towers  will  be  secured, 
at  any  joint,  if  the  obliquity  of  the  resultant  pressure  be  less 
than  the  angle  of  repose. 

Iron  and  timber  towers  are  to  be  treated,  each  as  a  whole, 
as  long  columns,  by  Gordon's  formula. 

If  the  anchorage  is  a  mass  of  masonry,  the  stabilities  of  po- 
sition and  friction  are  to  be  considered. 

Let  IV  =  weight  of  mass  and  q'  the  normal  distance  from  a 
vertical  line  through  its  centre  of  gravity  to  the  centre  of  fig- 
ure of  its  base.  Let  Tp  equal  the  tension  in  anchor  chains 
and  /  the  normal  distance  from  the  centre  of  pressure  to  its 
line  of  direction  ;  and  q  the  distance  from  the  centre  of  press- 
ure of  the  base  of  the  foundation  to  its  centre  of  figure. 

Then,  in  order  that  stability  of  position  may  be  secured  : 

TpPL^  lV{q  +  q) 
Stability  of  friction  is  secured   if  the  greatest  obliquity  of 


296 


SUSPEA^SIOiV  BRIDGES. 


the  resultant  pressure  on  any  section  (including  the  base)  is 
less  than  the  angle  of  repose  for  the  surfaces  in  contact. 

If  it  were  not  for  friction  between  the  anchor  chain  and  its 
supports,  on  the  circular  part  of  the  chain  (see  Fig.  15,  PI. 
XII.),  the  tension  would  be  the  same  throughout  its  whole 
length  ;  but  on  account  of  friction,  the  tension  diminishes  on 
the  circular  part,  from  link  to  link  downward,  according  to  the 
law  of  friction  between  cords  and  cylinders,  and  is,  therefore, 
the  least  at  the  bottom. 

The  diminution  of  tension  of  the  anchor  chain  is  computed 
by  this  formula : 


T 


TpEf' 


in  which  Tp  —  tension  of  anchor  chain  before  friction  takes 
effect ;  Tp  =  tension  of  any  point  below  the  first  point  of  sup- 
port ;  E  =  base  of  Napierian  system  of  logarithms  ;  f—  coef- 
ficient of  friction  ;  d  =  length  of  arc  considered.  The  anchor- 
age can  be  ruptured  only  by  the  breaking  of  chain,  or  bolt,  or 
plate,  or  pulling  out  the  whole  masonry.  The  probability  of 
the  latter  can  be  determined  by  comparing  the  tension  of  the 
chain,  at  the  upper  surface  of  the  masonry,  with  the  weight 
of  the  whole  masonry. 

Art.  64.* — Theory  of  the  Stiflfening  Tmss — Ends  Anchored — Continuous 
liOad — Single  Weight. 

It   has   been    seen    that  when  a  suspension    bridge    cable 
carries  a  load  covering  the  entire  span,  of  uniform  intensity 


Fig.  I. 


per  horizontal  unit,  its  centre  line  forms  a  parabolic  curve. 
When,  however,  such  a  cable  carries  an  isolated  weight,  or  a 
partially  uniform  load,  it  is  evident  that  the  centre  line  of  the 


*  This  and  the  following  two  Arts,  form  the  substance  of  a  paper  presented 
to  the  Pi  Eta  Scientific  Society,  in  June,  1879. 


THEORY  OF    THE   STIFFENING    TRUSS.  29/ 

cable  will  assume  a  form  different  from  that  of  the  preceding 
parabola,  unless  such  a  change  is  prevented  by  some  special 
device.     Such  a  special  device  is  the  stiffening  truss. 

The  objections  to  a  change  of  form  in  the  cable  of  a  sus- 
pension bridge  are  of  two  kinds.  Not  only  would  destructive 
undulations  result,  but,  also,  the  determination  of  stresses 
would  become  exceedingly  complicated  and  uncertain. 

Two  cases  may  arise :  the  stiffening  truss  may  be  securely 
anchored  at  its  ends ;  or  its  ends  may  simply  rest  upon  sup- 
ports and  be  free  to  rise,  in  which  case  there  can  be  no  nega- 
tive or  dozvnward  reaction. 

The  former  case  will  be  considered  first. 

It  is  desired  to  have  the  cable  retain,  for  all  positions  of  the 
moving  load,  the  same  parabolic  form.  Now,  it  has  already 
been  seen  that  such  a  result  can  be  attained  only  by  assuming 
a  uniform  pull  on  the  suspension  rods  from  end  to  end  of  the 
span.  Let  T  be  the  general  expression  for  this  uniform  pull 
for  any  suspension  rod,  and  let  /  be  its  intensity  per  unit  of 
span,  so  that  if/  be  the  panel  length  of  the  stiffening  truss. 
T  =  //.  Let  zv  be  the  weight  per  unit  of  span  of  the  fixed  load 
sustained  by  the  cables.  This  will,  of  course,  be  composed  of 
the  weights  of  the  truss,  suspension  rods  and  cable  or  cables. 
Let  w  be  the  moving  load  per  unit  of  span  ;  /  the  span  ;  R 
the  reaction  at  i?  ;  R  the  reaction  at  A,  and  let  the  moving 
load  pass  on  the  bridge  from  B. 

Also  let  x-^  be  the  distance  from  B  to  the  head  of  the  mov- 
ing load ;  the  latter  being  supposed  continuous  from  B. 

Since  all  the  forces  acting  on  the  stiffening  truss  are  vertical 
in  direction,  there  are  only  two  general  conditional  equations 
of  equilibrium,  and  those  simply  indicate  that  the  sum  of  all 
the  external  vertical  forces,  as  well  as  the  sum  of  the  moments 
of  the  same,  about  any  point,  must  be  zero. 

Those  two  equations  are  the  following : 

wl -{■  w' x^—tl  —  R —R  =  o  .     .     .     .     (i). 

(-W  +  w')^-t^-  Rx,+  U-  w)  ^^~  '""'^^  + 
'2  2  ^2 

/?'(/- .0  =  0.    .     .     (2). 


298 


SUSPE.VSWiV  BRIDGES. 


Eq.  (2)  can  be  at  once  written  by  taking  moments  about 
the  point  x^  at  the  head  of  the  moving  load. 

Eqs.  (i)  and  (2)  are  the  only  equations  of  condition  neces- 
sary for  equilibrium,  but  they  hold  three  unknown  quantities, 
i.  e.,  t,  R,  and  R  ;  hence,  any  one  of  those  three  quantities 
may  be  assumed  at  pleasure,  and  the  other  two  determined 
from  Eqs.  (i)  and  (2).  This  indetermination  simply  means 
that  unless  another  condition  be  imposed,  it  cannot  be  ascer- 
tained how  much  the  truss  will  carry  as  a  simple  truss,  and 
how  much  in  connection  with  the  cable. 

This  other  condition  is  virtually  the  following  :  the  stiffening 
truss  must  act  wholly  in  co7inection  with  the  cable,  and  carry  no 
load  zuhatever  as  an  ordinary  truss. 

The  direct  consequence  from  this  condition  of  the  problem 
is,  that  the  sum  of  all  the  uniform  upward  forces,  T  ^^  pt, 
must  be  equal  in  amount  to  the  sum  of  all  the  loads  of  the 
kinds  ^'  and  w .  But  the  line  of  action  of  the  resultant  of  the 
latter  is  not,  for  a  partial  moving  load,  the  line  of  action  of 
the  resultant  of  the  former:  consequent!},  the  truss  will  be 
subjected  to  the  action  of  a  couple.  In  order  that  equilib- 
rium may  be  assured,  therefore,  another  couple  of  equal  mo- 
ment, but  opposite  sign,  must  be  applied  to  the  truss ;  the 
forces  of  this  couple  must  act  at  the  extremities^  and  B,  and 
they  are  nothing  more  than  the  reactions  R  and  R .  From 
this  there  at  once  results  : 

R^  -  R. 
This  condition,  in  Eq.  (i),  gives: 

t=XV+   7C''^ (3). 

Eq.  (2)  gives : 

R  =  -R  =  '■^{^-■f).     .     .     .     (4). 

Eq.  (4)  shows  that  both  reactions,  R  and  R ,  are  zero  for 
Xi  —  /.  or  for  ic'  =  o. 


THEORY  OF   THE    STIFFENING    TRUSS. 


299 


It  is  also  seen  that  R  and  R  are  always  iiuincrical/)'  equal, 
but  liave  opposite  directions,  hence  R  is  a  doxvmvard  reaction, 
and  its  maximum  value  ivill  indicate  the  amount  of  anchorage 
required  at  each  end  of  the  truss. 

Using  Eq.  (4) : 

dR       IV       xu'x-L      IV  Xx  I 

dx-x  ~  2         2/  2/   ~"    '  '  *     ^       2  ■ 


Putting  Xi  =  "  in  Eq.  (4) : 


R  =  ^^ (5). 


Eq.  (5)  shows  the  maximum  value  of  (—  R)  and  gives  the 
amount  of  anchorage  required  at  either  end  of  the  truss ;  it 
aiso  shows  the  greatest  shear  to  be  provided  for  at  either  end 
of  the  truss. 

The  general  value  for  the  shear  at  any  section  of  the  por- 
tion of  the  truss  covered  by  the  moving  load  is : 


Or, 


S  =  R  +  tx  —  tvx  ~  w'x    ....     (6). 


w'xi       w'xi  Ixi         \     ,  ,  . 


This  value  of  5  shows  it  to  be  positive  near  the  end  of  the 
bridge  ;  it  then  decreases  as  x  increases,  passes  through  the 
value  zero,  and  then  increases  as  a  negative  quantity.  As  a 
negative  quantity  it  attains  its  maximum  value  for  x  =  x^;  it 
then  becomes : 


5.=  -'-^'(i--^)=^'    .     .     .     (8). 

Hence  the  two  reactions  at  the  ends  of  the  truss  and  the  shear 
at  the  head  of  the  moving  load  arc  always  numerically  equal. 


300  SUSPENSION  BRIDGES. 

I 

Eq.  (8),  consequently,  takes  its  maximum  value  for  x-^  =  ~ ,  and 

that  value  is  given  in  Eq.  (5)  ;  this  last  equation,  therefore, 
gives  the  maximum  shear  which  is  to  be  provided  for  at  the 
head  of  the  moving  load. 

Now  since  this  maximum  shear  is  to  be  provided  for  at 
both  ends  and  at  the  middle  of  the  truss,  it  would  probably 
be  advisable  in  all  ordinary  cases  to  design  all  the  web 
members  of  the  truss  to  be  of  uniform  size,  and  capable  of 
carrying  this  maximum  shear,  although  there  would  then  be 
a  little  waste  of  material  in  the  vicinity  of  the  quarter  points 
of  the  span  on  each  side  of  the  centre. 

This  supposes,  of  course,  that  the  chords  of  the  stiffening 
truss  are  parallel  and  horizontal.  If  the  chords  are  not 
parallel  the  amount  of  shear  carried  by  the  web  members 
will  depend  on  the  inclination  of  one  or  both  the  chords. 

For  all  values  of  x  and  x^,  for  the  portion  of  the  span  cov- 
ered by  the  moving  load,  the  total  shear  will  be  given  by 
Eq.  (7). 

For  the  portion  of  the  span  not  covered  by  the  moving 
load  the  general  value  for  the  shear  is  {measuring  x  from  A)  : 

^             _,,                           w'xi      w'  Xi       w'xiX  ,  . 

S=-  R'  +  wx-tx=~'-^^ -^      .    .    (9). 


This  expression  attains  its  greatest  values  for  x  =^  o  and 
X  =  I  —  x^.  In  the  first  case  the  shear  becomes  —  R',  and 
in  the  second  R.     These  results  show  nothing  new. 

Since  the  maximum  shear  in  a  simple  truss,  of  the  span  I 
and  uniform  loading  of  intensity  (w  +  w'),  is  i(w  +  'w')l,  it  is 
seen  that  the  maximum  shear  in  the  stiffening  truss  of  same 
span  is  only  one-fourth  of  that  due  to  the  moving  load  alone 
in  the  case  of  the  simple  truss. 

The  general   value  of  the  bending  moment  to  which  the 

truss  is  subjected,  for  the  portion   covered  by  the  moving 

load;  is: 

x^ 
M=  Rx  -{w  +  w'  -t)—      .     .     .     ( 10). 


THEORY  OF    THE    STIFFENING    TRUSS.  30 1 

Eq.  (10)  shows  that  if  x  =  -n,  the  bending  moment  is  equal 
to  zero.  Hence,  at  the  head  of  the  moving  load,  for  all  its 
positions,  there  is  a  section  of  contraflexure  or  no  bending,  and 
consequently  the  loaded  and  unloaded  portiotis  of  the  stiffening 
truss  are  each  in  the  condition  of  a  simple  beam  supported  only 
at  each  end,  and  loaded  uniformly  throughout  its  length. 

If  the  values  of  R  and  /  from  Eqs.  (3)  and  (4)  be  inserted 

in  Eq.  (10),  and  if  -^  be  put  equal  to  zero,  it  will  be  found 

that  the  bending  moment  has  its  maximum  value  for  x  ~  ■ — ; 
as  might  have  been  anticipated. 
Putting  X  —  —  in  Eq.  (10)  : 


2 


^^^  ....   ("). 


By  differentiating  in  respect  to  .I'l  it  will  be  found  that  M 
has  its  maximum  value  for  x-^  =  \l.  Denoting  this  value  of 
M  by  M^,  there  results : 

J/i  = (12). 

Eq.  (12)  shows  the  maximum  bending  moment  to  which 
any  loaded  portion  of  the  truss  can  be  subjected. 

If  a  simple  truss  supported  at  each  end  be  subjected  to  the 
action  of  a  uniform  load,  of  the  intensity  {%v  -f-  w'),  through- 
out its  entire  length,  the  greatest  bending  moment  will  be : 

M'  =  g-^  =  -^  (if  zv  =  o) ; 

.-.    J/o'  =  ^J/o'>^vW     .     .     •     (13)-* 
From  what  has  alread}'  been  shown  it  is  evident  that  the 

*  The  subscript  o  indicates  that  7z/  =  o  in  M'. 


302  SUSPE.VSION  BRIDGES. 

greatest  bending  moment   for  the  portion  of  the  truss  not 

covered  by  the  moving  load  will  occur  at  the  distance  ( '—) 

from  the  reaction  R .     The  general  value,  therefore,  for  the 
greatest  moment  for  that  portion  will  be : 

M=lR'{l-x,)  +  {i-zv)^-^^^    .    .    (14). 

Putting  the  differential  coefificient  of  M  in  respect  to  .Vj 
(after  inserting  the  values  of  R'  and  /)  equal  to  zero,  there 
results  : 


^j    —    -^l    — L-      V    9^^  9^ 

.-.     ^i  =  |/±-  =  /ori/      .     ._'  .     .      (15). 

The  latter  value  (^/)  gives  a  maximum,  and  inserted  in 
Eq.  (14): 

M,'  =  -  %'^'  =-M,=  -  \M^  (nearly)  .   .  (16). 
54 

Eqs.  (12)  and  (16)  show  that  the  greatest  bending  moments, 
to  which  the  stiffening  truss  is  subjected,  are  equal,  but  of  oppo- 
site kinds ;  and  it  is  seen  that  they  occur  when  one-third  or 
two-thirds  of  the  span  are  covered  by  the  moving  load.  The 
chords,  therefore,  of  the  stiffening  truss  must  be  designed  to 
resist  both  tension  and  compression. 

Eqs.  (10)  and  (14)  are  general  expressions  for  all  the  bend- 
ing moments  to  which  any  portion  of  the  truss  can  possibly 
be  subjected,  but  in  all  ordinary  cases  it  would  probably  be 
best  to  make  the  chords  uniform  in  section  (supposing  the 
depth  of  the  truss  to  be  constant)  from  end  to  end,  and 
capable  of  resisting  the  moments  given  by  Eqs.  (12)  and  (16). 

Eq.  (13)  shows  that  the  greatest  bending  moment  to  which 
a  stiffening  truss  can  be  subjected  is  only  \  of  that  found  in 
a  simple  truss  supported  at  each  end  and  loaded  with  a  uni- 


THEORY   OF   THE  STIFFENING    TRUSS. 


303 


form  load  equal  in  intensity  to  that  of  the  moving  load  on 
the  stiffening  truss. 

It  may  be  interesting  to  notice  that  the  resultant  load  on 
the  portion  -r^  of  the  truss  is  downward^  since   {zu  +  w')  > 

{t=.w-\-   ''    -^\  :    but    that   that    on    the    portion  (/  —  .r,)  is 

upward,  since  zv  <  /. 

If  the  bridge  is  traversed  by  a  single  concentrated  load  or 
weight,  W,  the  general  method  of  procedure  is  precisely  the 
same  as  before.  Let  the  weight  W  pass  on  the  bridge  from 
the  end  B,  in  the  figure,  and  let  x\  denote  its  distance  from 
that  point ;  also  measure  x  from  the  same  point. 

The  general  equations  of  condition  are : 

Rjr  R  +  tl-wl-W=^o     .     .     .     (17). 
- R'{l-x,)  +  {w  -  t)  ^"-"^^  -  {w -  t)  '^  +Rx,  =  o  .  (18). 


Eq.  (18)  is  written  at  once  by  taking  moments  about  the 
point  of  application  of  W. 

By  precisely  the  same  method  as  before,  there  may  be  found 
the  result,  R  =  -  R' . 

In  Eqs.  (17)  and  (18)  let  there  be  put  R  =  —  R',  then  there 
results : 

W 


x^ 


R=-R'^  ^^U-'y)    •    •    •     (20) 


When  Xi  =  -  ,  R  =  o  and  R'  =  o 


Eq.  (20)  shows  that  //-r  reactioji  nearest  the  weight  W  will 
always  be  positive,  or  upward ;  and  that  the  other  will  be  nega- 
tive, or  downward. 


304  SUSPENSION  BRIDGES. 

The  maximum  value  of  R  or  R'  is  found  (by  making  x\  —  o 

W 
in  Eq.  (20) )  to  be  — ,  and  that   ts  the  amount  of  anchorage 

required  for  the  weight  W,  alone,  at  eaeh  end  of  the  truss. 

The  point  of  appHcation  of  the  weight  IF  divides  the  span 
into  two  segments. 

The  general  value  of  the  shear  for  the  shorter  segment  is: 

S  =  tx-wx  +  R  =  ~x-\-  wU  -7-)  •  •  (21)- 
This  has  its  greatest  value  for  x  —  x-^\  it  then  becomes  : 

2 

Hence  S '  is  the  tmiform  maximum  upward  shear  that  must 
be  provided  for,  throughout  the  whole  length  of  the  truss. 

The  general  value  of  the  shear  in  the  longer  segment  of 
the  span  is : 

S=R'  -w{l-  x)+t{l-  x)^wU+Y  -^-^   .   .  (22). 

This  expression  attains  a  positive  maximum  for  x  =  x^, 
that  being  the  least  positive  value  of  x  admissible  ;  the  re- 
sulting value  of  5  is  ^  W,  which  shows  nothing  new. 

The  negative  maximum  for  ;r  =  /  is  simply  the  reaction  R'. 
Putting  5  =  o  in  Eq.  (22)  there  results  : 


X  =  Xi  -\ —  . 
^       2 

Hence  for  all  points  of  the  span  between  -  +  x^  and  /  there 

will  be  negative  or  downward  shear.  The  maximum  negative 
shear,  however,  to  be  provided  for,  is  shown  by  Eq.  (22)  to 
exist  in  one-half  of  the  truss  when  IF  rests  on  the  opposite  end ; 

or,  when  in  that  equation  x^  —  O  and  x  >  -.     The  negative 


THEORY   OF    THE   STIFFENING    TRUSS.  305 

or  downward  shear  to  be  provided  for  has,  then,  for  its  gen- 
eral expression  : 

S,=  -w{^-^-\)     ....     (23). 

Eq.  (23)  is  to  be  appHed  to  each  half  of  the  truss,  and  it  is 
also  seen  that  the  web  members  which  take  up  S^  should 
increase,  in  ultimate  resistance,  uniformly  from  the  centre  to 
the  ends  of  the  truss  ;  supposing  the  chords  to  be  parallel 
and  horizontal. 

In  many  cases,  however,  it  may  be  best  to  design  them  of 
uniform  dimensions  belonging  to  those  at  the  ends. 

The  bending  moment  for  any  point  of  the  smaller  segment 
of  the  truss  is  : 

M=Rx+{t-w)-^W{l--j-)x+  ^^     .    .    .    (24). 

Since  both  terms  of  this  moment  are  positive,  it  will  attain 
its  greatest  value  for  x  —  x^  ;  it  then  becomes  : 

J/,  =  -(.r,-^)      ....      (25). 

Eq.  (25)  gives  the  general  value  of  the  greatest  positive 
bending  moment  to  which  any  point  of  the  truss  will  be  sub- 
jected, for  which  case  x^  must  7iever  be  made  greater  than  \l. 

The  bending  moment  J/i  will  evidently  cause  compression 
in  the  upper  chord,  and  tension  in  the  lower.  • 

For  the  longer  segment  of  the  truss  the  general  value  of  the 
bending  moment  is : 


Or, 


M 
ao 


M=R'{l-x)^-  {t-  w)  ^ ^ 


=  _  j^(^i  _-ij(/_,^) +  ___v__i    ^    (26). 


3o6 


SUSPENSION  BRIDGES. 


There  is  evidently  a  point  of  contra-flexure  for  the  longer 
segment  of  the  truss,  for  if  the  second  member  of  Eq.  (26)  be 
put  equal  to  zero,  there  results  x  =  2^1.  Hence  all  the  por- 
tion I  —  2,Vi.  of  the  truss,  will  be  subjected  to  a  7iegative  bending 
moment  causing  tension  in  the  upper  chord  and  compression  in 
the  lower. 

In  Eq.  (26),  putting  -rj-, \  ==  0>  there  results  : 

^  =  ^1  +  - {^7)' 

This  value  of  x,  in  Eq.  (26)  gives  : 

Wl       Wx.       Wx? 

Eq.  (28)  gives  the  general  value  for  the  maximum  negative 
bending  moment  at  any  point  in  the  entire  truss,  for  any  position 
of  the  weight  W.    In  Eq.  (28),  it  is  to  be  remembered,  x^  must 

always  be  less  than  -  ;  also,  that  the  point  at  which  M'  exists 

will  be  given  by  the  value  of  x  in  Eq.  (27). 

The  formulae  for  a  continuous  load,  taken  in  connection 
with  those  for  a  single  weight,  will  give  all  the  circumstances 
of  bending  or  shearing  which  can  exist  with  any  condition  or 
position  of  loading. 

When  the  "  shear  "  has  been  determined  for  any  section, 
the  stress  in  the  web  member  which  is  to  carry  it  (if  the 
chords  are  parallel  and  horizontal)  will  be  found  at  once  by 
multiplying  that  "  shear  "  by  the  secant  of  inclination  of  the 
web  member  to  a  vertical  line.  If  there  are  two  or  more 
systems  of  triangulation  in  the  truss,  then  each  system  is  to 
be  treated  as  a  single  truss  in  the  usual  manner. 

If  desirable,  after  the  reactions  R  and  R  and  the  upward 
load  T  =  pt  are  known,  the  stresses  in  the  individual  members 
of  the  stiffening  truss  can  be  traced  as  in  the  case  of  an  ordi- 
nary truss  supported  at  each  end. 


THEORY   OF    THE    STIFFENING    TRUSS. 


307 


Art.  65. — Theory  of  the  Stiflfening  Truss — Ends  Free— Continuous  Load 

— Single  Weight. 

In  this  Article  the  notation  of  the  previous  one  will  be  con- 
tinued, and  the  same  figure  will  be  referred  to. 

The  case  of  a  continuous  load  will  first  be  treated,  and,  as 
before,  it  will  be  supposed  to  pass  on  the  bridge  from  B. 

Since  the  ends  are  not  anchored,  in  this  case  there  can  be 
no  negative  or  downward  reaction,  consequently  R'  will  be 
zero. 

As  before,  putting  the  sum  of  all  the  vertical  forces  acting 
on  the  truss  equal  to  zero,  and  taking  moments  about  the 
head  of  the  moving  load,  there  result  the  two  general  Equa- 
tions of  condition  : 

wl -V  wx\  — tl  —  R^o (i). 

(^  +  ^^,^  _  /)  ''^    -  (zf  -  t)  -^^^  -  Rx^  =  o  .   .    (2). 

Since  there  are  now  but  two  unknown  quantities,  R  and  /, 
the  problem  is  perfectly  determinate.     Eqs.  (i)  and  (2)  give  : 

^  =  «'''Vi(i-7-) (3) 


t  =  w+  —3 1^ (4) 

The  general  value  for  the  shear  at  any  section  of  the  truss, 
for  the  portion  covered  by  the  moving  load,  is 


S  =  R^[^t-w-  w)  X  =  IV  {x,  -  x)  -  w'  -Y[i  -  '^j    .    (5). 

Evidently  >S  has  its  maximum  positive  value  for  a:  =  o ;  its 
greatest  negative  value  for  x  =  Xi,  and  the  value  zero  for  x  = 
Jx^ 


3o8  SUSPENSION  BRIDGES. 

In  order  to  find  the  head  of  the  moving  load  for  that  po- 
sition which  makes  R  a  maximum,  let  -y—  be  put  equal  to 
zero. 

There  is  then  found  x\  —  — .  Hence  the  moving  load  cover- 
ing half  the  span  gives  the  maximum  reaction  R. 

I 
Placing  Xx^=  —   in  Eq.  (3),  the  greatest  value  of  R  becomes 

i?i  =  u''  /  -r-  4. 

In  order  to  determine  the  greatest  web  stresses  it  is  neces- 
sary to  find  the  greatest  shear  at  any  point  whose  abscissa  is 
X.  This  maximum  shear  at  once  results  by  placing  the  first 
derivative  of  5  in  respect  to  x^,  from  Eq.  (5),  equal  to  zero. 
That  operation  gives : 

2X^    (  X\  /2  • 

I r-  I     -    -7  I^    O    .-.    Xi    =    — , 


/     \  //  '        2(/-  ;r)* 

By  the  introduction  of  this  value  of  x-^  in  Eq.  (5),  the  great- 
est shear  for  any  section  located  by  x  becomes : 

i        /2 
max.  S  —  w'   \  —-, r  - 

U  (^  -  ^) 

It  is  clear  that  this  value  is  a  maximum  for  the  reason  that 

d^  S 

-j-^  is  a  negative  quantity  in  which  x^  does  not  appear. 

It  is  further  evident  that  max.  5  is  a  positive,  or  upward 
shear,  from  the  fact  which  was  observed  at  the  bottom  of 
the  preceding  page,  that  the  greatest  negative  shear  occurs 
at  the  head  of  the  moving  load.  By  making  x  =  x-^xn  Eq. 
(5)  that  greatest  negative  shear  becomes  : 

,  x^  (  Xi 

s,  =  -^  rV-7 

It  will  be  necessary  to  apply  max.  S  and  S ^  to  3.  half  of  the 
span,  regarding  the  shear  for  a  positive  direction  on  one  side  of 
the  centre  as  negative  for  the  other.  These  two  values  of  the 
shear  will  enable  all  the  greatest  web  stresses  to  be  determined. 

Since  R'  —  o,  the  whole  truss  will  be  subjected  to  bending 
moments  of  the  same  sign  ;  such  bending  moments,  in  fact,  as 


THEORY  OF    THE   STIFFENING   TRUSS.  309 

will  put  the  upper  chord  in  compression  and  the  lower  one  in 
tension. 

The  general  value  of  the  bending  moment  for  that  portion 
of  the  truss  covered  by  the  moving  load  is, 

J/=  Rx  -  {w  +  w  -  /)  —  , 

T,^  ,  W'X?X  7U'X^         W'X^X^  ,^. 

Ix 
Since  the  shear  S  is  zero  for  x  =    ,  "  ^ — ,  that  value  of  x  in 

/  +   Xy 

Eq.  (6)  will  give  the  maximum  value  of  M.     This  latter  is  : 
M^  =  wx^    f  ~  ^^  , (7). 

Putting  -j — =  o,  there  is  found, 
dxx 


;ri  =  -(-  I  ±  VS)  =  +  0.618/. 


The  absolute  maximum  bending  moment  exists,  therefore, 
when  the  moving  load  covers  0.618  of  the  span.  That  mo- 
ment has  for  its  value  : 

0.045 1  w7^  (nearly) (8). 

Ix 
If  ;ri  =  o.6i8/be  put  in  the  expression  ^  =  - — ? — ,  there  will 

/    "T"    -t  J 

result  X  =  0.382/,  nearly. 

Eq.  (6)  gives  the  general  value  of  the  bending  moment  for 
any  position  of  the  load,  but  it  would  probably  be  the  most 
convenient  to  make  the  moving  load  cover  0.618/,  and  design 
the  chords  for  the  distance  0.382/  from  each  end  to  resist  the 
bending  due  to  that  position  of  the  load,  and  then  design  all 
of  the  middle  2  (0.5  —0.382)/=  0.236/  to  resist  the  moment 


rjio  SUSPENSION-  BRIDGES. 

^iven  by  the  expression  (8).    By  this  arrangement  there  would 
be  a  little  surplus  of  material  at  the  middle  of  the  truss. 

If  a  single  weight  rests  upon  the  bridge,  the  two  general 
equations  of  equilibrium,  obtained  in  precisely  the  same  man- 
ner as  heretofore,  are  : 

wl^W-tl-R=o (9). 

(w  -  /)  ^  -  (w  -  t)  i!—^  -Rx,  =  o   .  .  (10). 

These  equations  then  give : 

2Wx,  f     . 


R=  w[\  -~^  .....     (12). 


It  Xi  =  - ,  i  =  w  +  ^-,  and  R  =  o. 
2  I 

The  general  values  of  the  shear  S,  and  moment  M,  are  the 
following : 

S  =  R  +  {t  -  %v)x, 
...     5  =  ^(i-^--)+^-i^.         .     (13). 

M=  Rx  -\-  U  -w)  — 
2 

...    M=w{.-^)x^^     .    .     (14). 

Eqs.  (13)  and  (14)  show,  since  Xi  must  not  be  taken  less 
than  X,  that  if  the  maximum  shear  and  bending  moment  are 
desired  for  any  section,  the  weight,  W,  must  be  placed  at  that 
section,  and  x  be  made  x^  in  those  equations. 


APPROXIAfATE   CHARACTER  OF  INVESTIGATIONS. 


311 


That  section  at  which  the  bending  moment  will  attain  its 
absolute  maximum  value  is  found  by  putting  x  —  x^  in  Eq. 
(14),  then  taking  the  first  differential  coefficient  of  M  in  re- 
spect to  ^'1,  equating  to  zero  and  solving.     There  results: 

x^^\l±\l=  \L 

This  value  in  Eq.  (14).  when  x  —  x^,  gives: 
M  =  3^  Wl. 

The  equations  for  the  continuous  and  single  moving  loads, 
used  in  combination,  will  give  moments  and  shears  for  any 
character  and  position  of  loading  whatever. 

The  general  observations  m  regard  to  finding  web  and 
chord  stresses,  at  the  close  of  the  last  Article,  apply  equally 
well  to  this  case. 

As  was  to  be  anticipated,  R  and  R'  in  the  two  preceding  Ar- 
ticles, are  independent  of  the  fixed  load  w. 


Art.    66. — ^Approximate    Character  of  the  Preceding  Investigations — De- 
flection  of  the  Tnass. 

In  the  two  preceding  Articles  it  has  been  virtually  assumed 
that  the  deflection  of  the  cable,  due  to  its  lengthening  under 
stress,  is  just  sufificient  to  allow  the  truss  to  take  the  deflec- 
tion due  to  the  loads  T  —  pt,  w',  and  W  xn  the  different  cases. 

Such,  however,  is  not  really  the  case. 

In  all  ordinary  cases,  the  cable  does  not  deflect  to  that  ex- 
tent. The  result  is,  as  is  evident,  that  the  truss  is  not  sub- 
jected to  the  amount  of  bending  assumed.  The  error, 
however,  is  only  a  small  one,  and  on  the  safe  side,  as  should 
be  the  case. 

It  has  been  found  by  experiment  that  if  the  ends  of  the 
truss  are  anchored,  the  stiffening  truss  will  be  subjected  to  a 
maximum  moment  equal  to  that  existing  in  an  ordinary  truss 
-supported  at  each  end,  with  the  same  span,  and  carrying  load 


312  SUSPENSION  BRIDGES. 

over  its  entire  length  of  about  one-eighth  the  intensity  of 
the  moving  load  on  the  suspension  bridge. 

This  same  case,  treated  analytically,  as  was  seen,  gave  about 
\  instead  of  \. 

Approximate  values  of  the  deflection  of  the  stiffening 
truss  can  be  found  by  the  ordinary  formulae  used  for  solid 
beams  in  the  subject  of  resistance  of  materials,  in  the  differ- 
ent cases,  where  /,  R,  R',  and  the  moving  load  are  known. 


CHAPTER   X. 

DETAILS   OF   CONSTRUCTION. 

Art.  67. — Classes  of  Bridges — Forms  of  Compression  Members— Chorda 

Continuous  or  Non-continuous. 

Regarding  the  systems  of  construction,  truss-bridge  struct- 
ures are  divided  into  two  classes  at  the  present  time,  i.  e., 
bridges  with  "  pin  connections "  and  bridges  with  "  riveted 
connections."  In  the  former  class  the  connection  of  web 
members  with  the  chords  or  with  each  other  is  made  by  a 
single  pin  only,  as  in  Fig,  3,  PL  III,  (the  figure  shows  simply 
the  upper  chord  and  tension  web  members  together  witn  one 
end  post).  The  pins  are  shown  at  1,2,  3,  4,  and  5,  where  the 
tension  members  join  the  chord.  In  the  latter  class  the  con- 
nections mentioned  are  made  by  means  of  rivets,  as  shown  in 
Fig.  6,  PI.  XI.  In  that  figure  ^  is  a  tension,  and  B  a  com- 
pression web  member,  while  C  is  a  portion  of  the  lower 
chord. 

Screw  connections  for  tension  web  members  and  simple 
abutting  connections  for  compression  ends  have  been  used, 
but  are  not  usually  employed  at  present. 

A  screw  connection  is  formed  by  passing  a  tension  member 
through  the  chord  at  one  of  the  joints,  and  placing  upon  the 
end  of  it  a  nut ;  this  method,  therefore,  can  only  be  conveni- 
ently used  when  the  tension  members  are  of  circular  cross- 
section. 

An  abutting  connection  for  a  compression  member  is 
formed  by  simply  abutting  either  end  against  the  chord, 
which  is  properly  formed  for  the  purpose  at  the  joint.  The 
end  of  the  post  or  strut  is  inserted  in  the  chord,  or  else  a 
projection  of  the  chord  passes  into  the  end  of  the  post ;  or, 
again,  some   simple   device  is  employed  for  the   purpose  of 

313 


314 


DETAILS  OF  CONSTRUCTION. 


keeping  the  ends  of  the  post  in  position,  and  for  nothing  else, 
as  the  entire  compressive  stress  in  the  post  or  strut  is  trans- 
mitted through  the  abutting  surfaces. 

Occasionally  screw,  abutting,  and  pin  connections  are  com- 
bined in  a  single  bridge. 

Without  re<.arding  systems  of  construction,  truss-bridge? 
are  divided  into : 

"  Deck  "  bridges,  /.  e.,  the  applied  load  is  on  the  chord  in 
compression. 

"  Through  "  bridges,  i.  e.,  the  applied  load  is  on  the  chord 
in  tension. 

"  Pony"  trusses  are  through  truss-bridges  when  the  trusses 
are  not  sufficiently  high  (or  deep)  to  need  overhead  cross- 
bracing,  and  they  are  seldom  put  up  for  spans  of  over  eighty 
feet,  although  there  are  examples  of  one  hundred  feet  (and 
even  more)  in  length. 

The  lateral  stability  of  such  long  pony  trusses,  however,  i:. 
very  precarious. 

Figs.  I  and  2  of  PI.  XI.  show  the  ordinary  forms  of  plate 
girder,  stringers  and  floor  beams,  with  plate  hangers  at  the 
ends  of  the  latter. 

The  various  forms  of  cross-sections  of  upper  chords  and 
posts,  and  compression  members  generally,  are  almost  innu- 
merable, and  subject  only  to  the  fancy  of  the  engineer  or 
builder.  The  principle  which  should  always  be  kept  in  view 
is  this :  That  the  material  should  be  as  far  as  possible  from 
the  neutral  axis.  Figs.  3,  4,  5,  and  6  of  PL  III.  show  methods 
of  building  up  the  upper  chord.  It  consists  of  riveting  plates 
to  a  pair  of  channel  bars,  or  to  a  pair  of  channel  bars  and  an 
X  beam. 

The  bars  or  beams  are  frequently  built  of  plates  and 
angles. 

The  blackened  portions  represent  sections. 
Chords  are  continuous  or  non-continuous  according  as  they 
are  built  up  in  a  continuous  manner  from  end  to  end,  or  built 
up  of  panels  abutting  against  each  other  at  the  panel  points. 
The  former  are  principally  used  at  present. 

Tension  members  are  always  of  rectangular  or  circular  sec- 


CUMULATIVE   STRESSES.  31$ 

tion.  Fig.  5,  PI.  XII.,  is  a  lower  chord  "eye-bar."  In  writing 
of  channel  bars,  I  beams,  angle-irons,  iron  bars  and  rods, 
they  are  indicated  as  follows:  C,  X,  L)  CD*  O;  that  is,  by 
skeletons  of  their  sections. 


Art.  68. — Cumulative  Stresses. 

Stresses  are  said  to  be  cumulative  in  any  part  of  a  struct- 
ure when  they  are  transmitted  through  that  part  to  other 
parts,  whose  whole  duty  is  to  sustain  them,  the  part  in  ques- 
tion being  subject  at  the  same  time  to  its  own  stress.  The 
member  in  which  the  stresses  are  cumulative  is,  therefore, 
overstrained  to  some  extent  in  some  one  or  more  portions  of 
it.  The  two  channel  bars  in  Fig.  3,  PI.  III.,  are  the  portions 
of  that  upper  chord  which  are  subjected  to  cumulative  stresses. 
If  C'C  is  supposed  to  be  the  centre  line  of  the  bridge,  then 
the  compressive  stress  in  the  chord  increases  as  C'C  is  ap- 
proached from  the  end,  in  consequence  of  the  components  in 
the  direction  of  the  centre  line  of  the  chord  of  the  stresses  in 
the  inclined  ties.  A,  B,  C,  etc.  This  increase  of  stress  is  pro- 
vided for  by  riveting  plates  to  the  upper  flanges  of  the  two 
Cs,  as  shown  in  the  figure.  It  is  evident  that  the  plates  do 
not  receive  the  stress  which  they  are  intended  to  bear,  except 
indirectly  through  the  Cs  and  the  rivets  which  connect  the 
latter  with  the  plates.  Now  since  the  Es  are  supposed  to 
have  their  own  share  of  direct  compressive  stress  to  sustain, 
it  is  plain  that  the  material  in  the  vicinity  of  the  front  of  the 
pin  (looking  from  the  centre  of  the  pin  toward  the  centre  of 
the  bridge)  is  subjected  to  a  much  greater  intensity  of  com- 
pressive stress  than  should  exist  in  the  structure.  This  re- 
lates only  to  the  material  in  front  of  the  pin,  and  that  is  the 
only  vicinity  in  which  cumulative  stresses  would  exist  if  the 
plates  could  be  so  securely  riveted  to  the  Cs  that  the  whole 
chord  could  be  depended  upon  to  act  as  one  piece.  In  prac- 
tice, however,  no  such  riveted  work  exists.  The  C^  must  in- 
evitably yield  to  some  extent  before  they  bear  sufficiently  on 
the  rivets  to  give  to  the  plates  their  proper  share  of  the 
stress.     The  result  is  that  not  only  the  material  in  front  of 


3l6  DETAILS  OF   CONSTRUCTION. 

the  pin  but  the  whole  of  the  Cs  are  overstrained  by  these 
cumulative  stresses.  The  only  remedy  is  to  so  proportion  the 
chord  that  those  parts  which  are  designed  to  sustain  stress 
shall  receive  it  immediately,  and  not  indirectly  through  some 
other  part. 

The  Fig.  5,  PI.  III.,  shows  a  method  of  accomplishing  this 
object.  The  plate  ab  is  a  light  one  riveted  to  the  top  flanges 
of  the  Cs,  and  extends  throughout  the  whole  length  of  the 
chord.  The  increase  of  the  areas  of  cross-sections  are  ob- 
tained by  riveting  plates  to  the  flat  sides  of  the  Qs,  and  by 
adding  an  X>  if  necessary,  as  shown.  The  parts  of  the  chord 
thus  receive  stresses  immediately  from  the  pins,  and  cumula- 
tive stresses  are  obviated. 

It  is  also  evident  that  if  the  stresses  are  applied  to  the  cen- 
tres of  gravity  of  the  cross-sections,  or  parts  of  the  cross-sec- 
tion, no  cumulative  stresses  will  exist. 

Cumulative  stresses  are  as  liable  to  occur  in  riveted  connec- 
tions as  in  pin  connections  ;  in  fact,  more  so.  It  may  be  said 
to  be  impracticable  to  so  construct  riveted  work  that  cumula- 
tive stresses  will  not  exist,  and  in  this  respect  pin  connections 
have  the  advantage  of  riveted  connections.  Fig.  6,  PI.  XI., 
illustrates  the  matter  for  a  riveted  chord.  The  plate  C  is 
common  to  the  whole  chord,  and  all  web  members  are  riveted 
to  it,  as  shown  by  A  and  B,  so  that  before  the  rivets  can  take 
their  share  of  the  stress  in  transferring  it  to  the  plate  and  Ls 
ED,  it  (the  plate  C)  will  necessarily  yield  to  such  an  extent 
that  cumulative  stresses  will  exist  throughout  its  whole  length 
to  a  greater  or  less  degree. 

Art.  69. — Direct  Stress  Combined  with  Bending  in  Chords. 

If  direct  stress  is  not  applied  to  the  centres  of  gravity  of 
the  ends  of  a  piece  subjected  to  compression,  it  is  clear  that 
bending  must  take  place. 

In  Figs.  3  and  4  of  PI.  III.,  the  horizontal  components  of 
the  oblique  forces  in  the  ties  A,  B,  etc.,  do  not  act  through 
the  centres  of  gravity  of  the  sections  of  the  chord,  hence 
there  must  be  a  bending  in  those  chords.     If  the  chords  were 


DIRECT  STRESS    WITH  BENDING  IN  CHORDS.  317 

perfectly  straight,  and  if  the  centres  of  the  pin-holes  were  all 
at  the  same  distance  from  the  centres  of  gravity  of  the  dif- 
ferent cross-sections,  as  well  as  in  the  same  straight  line,  then 
the  total  direct  stress  to  which  the  chord  is  subject  at  any 
section  would  produce  bending  at  that  section,  and  the  lever, 
arm  would  be  the  same  for  all  sections.  Camber  and  deflec- 
tion from  loading,  however,  so  complicate  the  matter  that  it 
is  quite  impossible  to  make  even  a  satisfactory  approximate 
computation  of  the  chord  bending  arising  from  this  cause. 
All  chords  in  compression,  therefore,  should  be  so  designed 
that  the  axes  of  the  pins  may  traverse  the  centres  of  gravity 
of  their  sections,  even  though  the  ties  rest  directly  on  the 
upper  chord.  It  is  clear  that  when  this  bending  exists,  the 
proper  distribution  of  direct  stress  is  greatly  disturbed,  though 
to  an  indeterminate  extent  ;  and  is,  except  in  most  rare  cases, 
a  very  faulty  construction. 

If  it  be  supposed  that  the  total  direct  stress  in  the  chord 
acts  as  if  the  latter  were  perfectly  straight,  so  that  it  all 
produces  flexure ;  and  if  it  then  be  supposed  that  the  in- 
crement only,  at  each  pin  or  panel  point  produces  flexure  in 
the  adjacent  panel ;  it  is  evident  that  the  first  supposition 
will  make  the  flexure  the  greatest  possible,  while  the  second 
will  make  it  the  least  possible. 

It  is  farther  evident  that  if  cumulative  stresses  occur,  flex- 
ure must  necessarily  exist,  for  the  simple  reason  that  the  di- 
rect stress  is  not  uniformly  distributed  over  the  cross-section 
of  the  chord. 

Although  this  flexure  is  indeterminate  in  amount  and  shows 
a  faulty  design,  the  attempt  to  utilize  it  has  sometimes  been 
made,  and  the  analysis  on  which  the  practice  was  based  will 
now  be  given,  it  being  premised  that  the  ties  are  supposed 
to  rest  directly  on  the  chords,  as  shown  in  Figs.  4  of  Pis.  III. 
and  I. 

Suppose,  in  PL  III.,  Fig.  6  to  be  an  enlarged  cross-section 
of  the  chord  in  Fig.  4,  and  let  fg  pass  through  the  centre  of 
gravity  of  the  cross-section,  being  parallel  to  ab  and  cd' ; 
then,  since  the  increment  of  the  chord  stress  transmitted 
through  the  pin  from  the  ties  AA  is  applied  to  the  cross-sec- 


318 


DETAILS   OF   CONSTRUCTION. 


tion  of  the  chord  at  a  distance  h  below  the  centre  of  gravity; 
there  will  be  an  excess  over  the  uniform  intensity  of  stress  in 
the  cross-section  at  the  lower  side  EE,  and  a  deficiency  at  the 
upper  side  ab. 

This  excess  or  deficiency  (the  same  in  amount,  of  course, 
for  certain  sections  only)  is  found  in  a  very  simple  manner,  as 
follows : 

Let  P  be  the  increment  of  direct  compressive  stress  given 
to  the  chord  by  the  ties  AA,  and  let  P^  be  the  total  direct 
compressive  stress  in  the  section.  Put  .S"  for  the  area  of  the: 
cross-section. 

Now  the  variation  of  the  intensity  of  stress  from  the  mean 
is  due  to  the  moment  Ph,  and  since  this  moment  is  constant 
for  all  points  between  any  two  pins,  as  i — 2  or  2 — 3,  Fig.  4, 
PI.  III.,  the  variation  in  intensity  is  also  constant  between^ 
these  points. 

RT 
The  moment  Ph  —  —-t-  ,  in  which  rt^  equals  the   distance 
"1 
from  /g-  to  ££,  and  R  the  intensity  of  stress  at  EE  due  to- 
bending,  gives 

The  intensity  of  stress  at  ad  due  to  bending  is,  of  course,, 
equal  to 

Now  the  total  intensity  of  stress  at  EE,  Fig.  6,  PI.  III.,  is- 

P  P  d  —  d 

equal  to  -7^  +  i?;  and  that  at  ab,  -^  —  R — -:, — ^;  the  inten- 

p 
sity  at  7^=-^,  whatever  may  be  the  figure  of  the  cross-sec- 
tion.    The  variation  of  intensity  at  any  point  in  the  section 
may  be  easily  found  from  7^  by  a  simple  proportion,  and  the 

p 
total  intensity  by  adding  that  to  ~. 


DIRECT  STRESS    WITH  BENDING  IN  CHORDS.  319 

As  a  first  case,  let  the  chord  be  a  non-continuous  one,  so 
that  each  panel,  so  far  as  the  panel  moving  load  is  concerned, 
is  a  simple  beam  supported  at  each  end. 

If  the  load  rests  on  the  upper  chord  immediately,  as  shown 
in  Fig.  4,  PI.  III.,  it  will  produce  tension  at  the  lower  side  of 
the  chord  and  compression  at  the  upper  by  simple  flexure,  an 
opposite  tendency  to  that  exerted  by  the  moment  Ph. 

The  moments  due  to  the  moving  load  on  any  panel  vary 
(that  is,  increase)  from  the  joints  to  the  middle  point  of  the 
panel,  where,  of  course,  the  moment  is  maximum.  Denote 
by  R'  the  greatest  intensity  of  the  tensile  stress  caused  by 
the  moment  of  the  moving  load,  then 


^  _  d^'2wx^ 


I 


in  which  'S.WX'^  expresses  the  greatest  moment  of  the  applied 
load. 

For  a  uniform  load  : 

^WXx  -  — g-  , 

and  it  exists  at  the  centre  of  the  panel.  Now,  ordinarily,  the 
chord  would  be  required  to  resist  the  bending  moment  ex- 
pressed by  '^wx,  but  h  may  be  so  chosen  that  for  its  max- 
imum value  R  —  R',  and  then  no  extra  metal  will  be  required 
on  account  of  the  flexure  produced  by  the  moving  load.  This 
yalue  of  h  is  found  as  follows  :  put 

Phd]^  _  d{2wx-^  _  2wxj 

In  all  ordinary  cases  of  uniform  load 

2wx^  =  —^  > 


220  DETAILS  OF  CONSTRUCTION. 

in  which  w  is  the  intensity  of  uniform  load. 

•••     ^^  =  ^- 

The  value  of  h,  therefore,  is  independent  of  the  form  of 
cross-sections.  \i  R  <,  R',  additional  material  will  be  needed 
in  order  to  prevent  an  excess  of  compressive  stress  at  the 
upper  part  of  the  chord,  and  a  deficiency  at  the  lower  side. 
When  R  >  R' ,  there  is  an  excess  of  compressive  stress  at  EE, 
Fig.  6,  PI.  III. 

When  the  lower  chord  sustains  the  moving  load  directly, 
as  in  Fig.  4,  PI.  XL,  the  only  change  arises  from  this :  That 
it  is  in  tension  instead  of  compression,  and  h  is  measured 
above  the  centre  of  gravity  of  the  cross-section,  instead  of 
below  it  ;  also,  the  section  is  rectangular. 

In  the  case  of  the  lower  chord,  however,  if  the  stress  in  one 
panel  is  given  to  the  adjacent  one  through  the  medium  of  a 
pin,  then  the  total  stress  in  the  panel  under  consideration 
must  be  put  for  P  in  the  formulae  above.  This  must  also  be 
done  in  every  case  where  the  total  chord  stress  produces 
bending.  The  chord  stress  used  may  be  taken  as  the  max- 
imum (that  which  exists  when  the  moving  load  covers  the 
whole  truss),  for  in  all  other  cases  there  is  a  surplus  of  ma- 
terial with  which  to  resist  the  bending. 

This  method  of  neutralizing  the  flexure  produced  by  the 
direct  application  of  the  moving  load  to  the  chords  is  very 
unsatisfactory  in  many  ways  and  should  never  be  used.  At 
all  places  in  the  panel  except  the  centre  the  metal  is  still 
over-strained  by^  the  flexure  due  to  its  own  stress,  and  in  the 
vicinity  of  the  panel  points  this  condition  exists  to  a  very 
serious  extent.  When  this  consideration  is  coupled  with  the 
great  uncertainty  attached  to  the  hypothesis  on  which  the 
analysis  is  based,  the  unsatisfactory  character  of  the  method 
is  sufficiently  evident  to  effect  its  exclusion  from  the  best 
practice. 

If  the  chord  is  continuous,  the  objections  to  the  method  al- 
ready mentioned    gather   considerably  increased   force.     At 


DIRECT  STRESS    WITH  BENDING  IN  CHORDS.         321 

and  near  the  ends  of  the  panels  the  fixed  and  moving  load 
produce  flexure  in  the  same  direction  as  the  direct  chord 
stress,  and  to  twice  the  amount  of  that  at  the  centre.  It  is 
not  necessary,  therefore,  to  consider  this  case  farther. 

A  considerable  saving  of  material  can  be  effected  by  plac- 
ing the  ties  directly  on  the  upper  chord  of  a  deck  bridge,  and 
with  a  proper  design  it  in  no  manner  conflicts  with  the  best 
practice.  In  all  cases  the  axis  of  the  pm  should  traverse  the 
centre  of  gravity  of  the  chord  section,  or  as  nearly  so  as  practi- 
cable, in  order,  if  possible,  to  eliminate  all  flexure  due  to  the 
direct  chord  stress.  The  chord  section  should  then  be  so 
formed  that  the  combined  stresses  due  to  flexure  in  the  ex- 
terior fibres,  and  the  direct  chord  stress  shall  at  no  point  ex- 
ceed a  proper  value  per  square  unit.  This  value  may  be 
taken  at  8,000  to  9,000  pounds  per  square  inch  for  wrought- 
iron  upper  chords,  or  10,000  to  11,000  for  mild  steel  members 
of  the  same  kind.  These  values  may  be  taken  comparatively 
high  for  the  reason  that  an  indefinitely  small  portion  only  of 
the  material  is  subjected  to  these  intensities,  and  that  small 
portion  is  well  supported  against  fatigue  by  the  material 
about  it,  which  is  considerably  understrained. 

If  the  notation  previously  used  in  this  Art.  be  still  main- 
tained, the  maximum  external  moment  '^wx-^  will  develop  in 
the  most  remote  fibres  at  the  distance  d^  from  the  neutral 
axis  the  intensity. 

R^d^^P ^^^^ 

If,  on  the  other  hand,  P^  is  the  total  direct  stress  in  the 
chord  and  5  the  area  of  cross  section,  while/  is  the  greatest 
allowable  combined  stress,  then  there  will  result : 

di^zux.      Pi  ,  . 

^  =  ^T^+S •    •    ■    (^)- 

Eq.  (2)  shows  that  in  the  most  ef^cient  design  the  moment 
of  inertia  /of  the  section  must  be  the  greatest  possible.  At 
the  same  time  considerations  affecting  the  joint  details  render 


322  DETAILS  OF  CONSTRUCTION. 

it  advisable  that  the  centre  of  gravity  should  lie  not  far  from 
the  mid-depth  of  the  section.  These  two  conditions  are  ful- 
filled by  placing  large  quantities  of  the  material,  and  as 
nearly  as  possible  in  equal  amounts,  at  the  top  and  bottom 
of  the  chords,  as  shown  in  Fig.  i8  of  PI.  XII.  The  cover  plate 
be  and  angles  dd  are  made  as  light  as  the  circumstances  of 
proper  design  will  permit,  but  the  angles  aa  are  made  as 
heavy  as  possible.  An  unequal-legged  angle  is  a  very  good 
one  {ox  aa  with  the  longest  leg  horizontal,  and  it  is  sometimes 
necessary  to  rivet  a  narrow  plate  to  those  horizontal  legs  in 
order  to  properly  balance  the  section.  The  centre  of  gravity 
XxViO.  fg  will  usually  lie  a  little  above  the  centre  of  figure. 

As  the  centre  of  the  span  is  approached  from  the  end  the 
chord  section  must  be  rapidly  increased  but  in  no  case  should 
that  increase  be  made  by  tkickenifig  the  cover  plates  or  increas- 
ing their  number,  as  such  an  operation  inevitably  means 
cumulative  stresses  or  flexure  by  direct  stress.  In  rare  cases 
it  may  be  admissible  to  slightly  thicken  a  cover  plate,  if  there 
is  but  one,  but,  as  a  rule,  Fig.  6  of  PI.  III.,  shows  a  design  to 
be  carefully  avoided.  All  increase  of  section  should  be  ob- 
tained by  thickening  the  side  or  web  plates,  or  increasing 
their  number  ;  or,  again,  by  increasing  the  angles,  or,  finally, 
by  introducing  an  interior  eye-beam,  as  shown  in  Fig.  5, 
PI.  III. 

If  the  chord  is  non-continuous,  '^wx^  is  simply  the  bending 
for  a  span  equal  in  length  to  a  panel  and  due  to  the  track 
load,  own  weight  and  superimposed  moving  load,  and  is  easily 
determined.  If  the  chord  is  continuous,  on  the  contrary,  the 
analysis  for  the  moving  load  bending  is  not  simple.  The 
total  bending  for  this  case,  however,  may  properly  and  safely 
be  taken  at  three-fourths  its  value  for  a  non-continuous 
chord. 

The  upper  chord  section  required  in  the  case  of  combined 
bending  and  direct  stress  is  readily  found  by  the  aid  of  Eq. 
(2).  The  radius  of  gyration  r  can  be  easily  and  with  suf- 
ficient accuracy  predetermined  ;  so  that  S-r  can  be  put  for  / 
in  that  equation.  After  that  substitution  is  made,  there  at 
once  results  : 


RIVETED  JOIXTS.  323 

S  =  i(^'  +  /',) (3). 

This  is  a  very  convenient  formula  for  practical  use. 

Art.  70.— Riveted  Joints  and  Pressure  on  Rivets. 

In  riveted  bridge  work  the  pitch  of  rivets  {i.  e.,  the  dis- 
tance from  centre  to  centre)  should  not  be  less  than  three 
diameters,  although  it  sometimes  is  ;  if  possible  it  should  be 
from  four  to  eight  diameters  of  the  rivet,  provided  that  value 
does  not  exceed  about  fourteen  or  sixteen  times  the  plate 
thickness.  The  diameter  of  the  rivet  is  determined  by  the 
amount  of  stress  which  the  joint  is  to  carry,  so  that  the  in- 
tensity of  pressure  against  the  surface  of  the  rivet  in  contact 
Avith  the  plate  shall  not  exceed  a  given  value.  If  the  rivet 
and  hole  were  in  ideally  perfect  contact,  this  intensity  could 
easily  be  found,  having  given  the  amount  of  stress  which  the 
rivet  is  to  carry.  But  such  is  never  the  case.  The  only  re- 
sort left,  therefore,  is  to  assume  that  the  rivet  does  fit  per- 
fectly, and  fix  a  low  enough  value  for  the  intensity  of  pressure 
against  its  surface  to  make  the  joint  safe. 

In  Fig.  I,  PI.  XII.,  suppose  EF to  be  a  part  of  a  plate  in 
which  is  drilled  or  punched  the  rivet-hole  ADBK,  and  sup- 
pose the  stress  to  be  exerted  on  the  plate  in  the  direction  of 
the  arrow  at  K,  then  the  surface  of  contact  between  the  plate 
and  rivet  will  be  projected  in  ADB ;  contact  will  not  take 
place  throughout  the  whole  semi-circumference  when  the 
plate  is  not  subjected  to  stress,  unless  the  rivet  fits  the  hole 
with  absolute  accuracy. 

Since  all  material  is  elastic  to  some  degree,  there  will  be  a 
surface  of  contact  when  the  plate  is  subject  to  stress,  even 
when  the  rivet  does  not  accurately  fit  the  hole,  and  this  sur- 
face will  evidently  increase  with  the  stress  in  the  plate. 

But  suppose  that  the  rivet  fits  the  hole  exactly,  then  the 
pressure  on  the  surface  of  contact,  ADB,  will  be  of  uniform 
intensity,  and  the  case  will  be  similar  to  that  of  fluid  pressure 
on  a  cylinder.  Let/  denote  this  intensity  whose  direction  is 
normal  to  ADB  at  every  point  of  it  (friction  is  omitted  from 


324  DETAILS   OF   CONSTRUCTION. 

consideration),  then  the  total  pressure  in  the  direction  of  the 
arrow  at  K  exerted  by  the  plate  on  the  rivet  for  each  unit  of 
length  of  the  latter  is  equal  to  pAB. 

Let  t  be  the  thickness  of  the  plate,  as  shown,  and  put  d  for 
the  diameter  AB,  then  the  total  pressure  against  the  rivet  in 
the  direction  of  the  arrow  is 

P^ptd. 

The  quantity/  is  the  greatest  mean  value  of  the  intensity 
of  compressive  stress  which  it  is  desirable  to  put  upon  the 
material  under  the  given  circumstances.  It  is  usually  taken 
as  high  as  12,000  lbs.  per  square  inch,  although  10,000  is 
a  safer  value.  Of  course,  the  actual  maximum  value  of  the 
intensity  immediately  in  front  of  the  centre,  (T,  is  much 
greater  than  either  10,000  lbs.  or  12,000  lbs.  If  Z'  is  the 
amount  of  stress  which  the  joint  is  required  to  carry,  then 
the  number  of  rivets,  so  far  as  the  previous  consideration  is 

concerned,  is  equal  to 

T' 

■ 7  =  n. 

ptd 

The  riveted  joint  itself,  as  shown  in  Fig.  5,  PI.  XL,  may 
now  be  examined.  The  distance  c  should  be  at  least  23^ 
diameters  of  the  rivet.  By  the  arrangement  of  the  rivets 
shown,  when  the  pitch  is  from  four  to  eight  diameters,  the 
strength  of  the  plate  of  the  width  w  will  only  be  decreased 
by  about  the  amount  of  metal  taken  out  in  one  rivet-hole, 
although  experiments  to  settle  this  point  definitely  are 
wanting. 

After  having  determined  the  pitch  and  distance  of  r,  as 
above,  there  are  five  methods  of  rupture  of  the  joint  only 
which  need  serious  attention.  These  five  are:  (i)  tearing  of 
the  plate  though  the  rivet-hole  E,  (2)  tearing  of  the  cover- 
plates  through  the  rivet-holes  at  the  middle  of  the  joint,  two 
in  the  figure,  (3)  shearing  of  the  rivets,  (4)  and  (5)  rupture  by 
compression  at  the  surface  of  the  contact  between  the  rivets 
and  the  plates.  The  safe  shearing  stress  to  which  rivets  are 
subjected  in  bridge  structures  is  usually  taken  at  7,500  lbs. 
This  gives  a  safety  factor  of  from  5  to  6. 


RIVETED    CONNECTIONS.  325 

Put  5  for  the  intensity  of  the  maximum  safe  shearing  stress 
on  rivets  (7,500  lbs.  for  wrought  iron),/  for  the  intensity  of 
the  maximum  compressive  stress  (10,000  to  12,000  for  wrought 
iron),  and  7"  for  the  maximum  working  tensile  stress;  also,  ri 
for  the  number  of  rivets  on  the  line  through  the  middle  of  the 
joint  (two  in  the  figure).  Let  t  and  /  '  represent  the  thickness 
of  the  plate  and  covers  as  shown.  Then  equal  liability  to 
rupture  in  the  five  ways  mentioned  is  expressed  as  follows : 

Tt  (zu  —  d)  =  2  TV'  {w—n'd)  = .  2  .  S  =  ntdp  = 

4 

2nt'dp  =^  T'. 

It  almost  always  happens  that  these  quantities  are  not  each 
equal  to  T',  but  none  of  these  should  be  less.  If  only  one 
cover-plate  is  used,  the  2Tt '  should  be  replaced  by  Tt ',  2S  by 
S,  and  27it'  by  nt'.  The  form  of  this  Equation  of  condition 
may  be  somewhat  changed  by  piling  of  the  plates,  etc.,  but  it 
will  remain  essentially  the  same,  and  serves  to  illustrate  the 
principle  which  must  govern  in  all  cases. 

We  see,  therefore,  that  in  obtaining  the  amount  of  pressure 
which  should  be  put  upon  a  rivet,/  ought  to  be  multiplied  by 
its  diameter,  and  not  by  the  semi-circumference  ADB,  Fig.  i. 

Art.  71. — Riveted  Connections  between  Web  Members  and  Chords. 

When  web  members,  as  A  and  B,  Fig.  6,  PI.  XL,  are  riv- 
eted to  the  chord,  the  centre  line  (/.  r.,  the  line  joining  the 
centres  of  gravity  of  the  sections  of  the  members)  should  pass 
through  the  centre  of  gravity  of  a  system  of  points  situated  at 
the  centres  of  the  rivet-holes  ;  otherwise  the  intensity  of  stress 
in  any  section  of  the  member  will  not  be  uniform,  and  it  (the 
web  member)  will  be  subjected  to  flexure.  It  is  supposed, 
of  course,  that  each  rivet  carries  the  same  amount  of  stress, 
which,  however,  is  probably  seldom  true,  but  it  is  the  best 
assumption  that  can  be  made.  Fig.  6  represents  a  proper  dis- 
tribution of  rivets  in  reference  to  the  centre  lines  Ac  and  Be. 

It  will  be  observed  that  the  strut,  composed  of  two  un- 
equal legged  angles,  has  its  connection  with  the  chord 
through  both  legs  by  means  of  the  angle  lugs.     This  should 


326  DETAILS   OF   CONSTRUCTION. 

always  be  done  in  similar  cases,  for  in  no  other  way  can  an 
angle-brace  develop  its  full  strength.  The  practice  of  rivet- 
ing single  legs,  only,  of  angle-braces  to  chords  is  highly  ob- 
jectionable, for  the  reason  that  the  actual  resistance  of  the 
brace  is  far  below  the  nominal. 

When  three  or  more  pieces  are  riveted  together  at  the 
same  joint,  all  the  centre  lines  of  stress  should  intersect  at 
one  point,  if  flexure  is  to  be  avoided.  It  is  frequently  im- 
practicable to  do  this  in  riveted  connections,  and  the  imprac- 
ticability constitutes  a  serious  objection  to  that  character  of 
work. 

If  T  is  the  total  tension  in  the  member  A  of  Fig.  6,  PI.  XL, 
and  C  the  compression  in  B,  then  there  will  be  developed  at 
a  the  bending  moment  : 

T  ^  ac  sin  6  ; 

and  at  b  the  bending  moment : 

C  ^  be  sin  d  ; 

it  being  supposed  that  ac,  be,  and  ab  are  the  centre  lines  of 
stress  of  the  two  members  and  chord. 

It  is  very  true  that  the  metal  is  well  supported  in  the 
vicinity  of  the  joint,  but  unless  provision  is  made  for  the  flex- 
ure, as  shown  in  Art.  69,  which  is  seldom  or  never  the  case, 
some  of  the  metal  will  be  over-strained.  Hence,  this  flexure 
should  always  be  made  a  minimum,  and  reduced  to  zero  if 
possible. 

In  pin  connections  this  bending  at  the  joints  is.  of  course, 
entirely  obviated  when  all  centre  lines  of  stress  intersect  at 
the  centre  of  the  pin. 

Art.   72. — Floor-Beams  and  Stringers. — Plate  Girders. 

The  load  applied  to  a  bridge  rests  immediately  on  the  floor- 
beams,  generally  speaking,  and  is  transferred  through  them 
to  the  joints  of  the  truss.  In  railway  bridges  the  track  and 
ties  lie  on  stringers,  which  rest  on  the  floor-beams.  There  are 
two  or  more  stringers  for  each  track. 

In  highway  bridges,  the  floor-beams  support  stringers,  say 


FLOOR-BEAMS  AND    STRINGERS.  32/ 

two  feet  apart  (sometimes  less  and  sometimes  a  little  more), 
running  parallel  to  the  centre  line  of  the  bridge,  which  carry 
the  floor. 

In  railway  bridges,  the  moving  load  is  applied  to  the 
beams  at  the  ends  of  the  stringers,  but  in  highway  bridges 
the  greatest  moving  load,  for  which  the  beam  is  to  be  de- 
signed, may  be  taken  as  uniformly  distributed  over  the  entire 
length  of  the  beam,  or  rather  that  part  of  it  between  the 
points  of  support. 

Floor-beams  should  always  be  supported  at  the  ends  by  a 
single  hanger,  or  by  some  equivalent  arrangement,  which  rests 
at  the  centre  of  the  pin,  or  centre  of  the  chord  in  riveted  con- 
nections. Double  hangers  may  be  made  tolerable  by  some 
equalizing  device,  usually  of  an  expansive  character,  but  as  a 
rule  they  cannot  be  too  strongly  condemned,  for  in  such  cases 
the  deflection  of  the  beam  will  throw  the  greater  part  or  all 
of  the  weight  of  the  beam  and  its  load  on  the  inner  hangers. 
The  result  will  be  not  only  a  great  overstraining  of  the  latter, 
but  a  prejudicial  redistribution  of  stresses  in  both  web  mem- 
bers and  chords.  The  excessive  load  on  the  inner  hangers 
will  cause  an  overstrain  in  the  inner  tension  braces  which  will 
extend  to  the  inner  lower  chord  members,  and  even  to  the 
posts  and  upper  chord. 

Floor-beams  are  frequently  built  into  vertical  posts.  In 
such  cases  an  essentially  central  bearing  on  the  pin  should 
be  provided. 

Plate  girder  stringers  for  railway  bridges  arc  either  sup- 
ported in  between  the  floor-beams,  or  partially  so  and  par- 
tially above,  or  are  supported  wholly  on  the  top  of  the  floor- 
beams.  With  proper  designing  there  is  little  difference  in 
cost  in  the  various  methods.  The  first,  however,  is  far  prefer- 
able, for  the  reason  that  it  gives  the  greatest  stifl'ness  to  the 
floor  system,  other  things  being  equal. 

The  depths  of  plate  girder  stringers  is  usually  found  be- 
tween one-ninth  and  one-twelfth  of  their  spans,  i.  e.,  the 
panel  lengths.  The  floor-beam  depth  should  be  as  great  as 
economical  considerations  will  permit,  in  order  that  the  de- 
flections may  be  the  smallest  possible. 


328  DETAILS  OF   CONSTRUCTION. 

The  webs  of  stringers  and  floor-beams  slightly  aid  resistance 
to  flexure,  but  rivets  in  stiff'eners  and  splice  plates,  if  such  ex- 
ist, decrease  this  resistance  to  some  extent.  Hence,  it  is  the 
best  practice  to  disregard  the  resistance  of  the  web  to  flexure, 
and  to  assume  that  it  resists  the  shear  only.  This  is  the  more 
advisable  when  it  is  remembered  that  the  rivets,  in  giving 
stress  to  the  flanges,  produce  a  flexure  in  the  latter  which  is 
always  neglected.  This  flexure  arises  from  the  fact  that  the 
rivet  holes  never  pass  through  the  centre  of  gravity  of  the 
flange  angle  section. 

The  true  depth  of  a  plate  girder  is  the  vertical  depth  be- 
tween the  rivet  hole  centres,  but  by  a  curious  confusion 
between  rolled  and  built  sections,  it  is  commonly  taken  as  the 
vertical  distance  between  the  centres  of  gravity  of  the  flange 
angles. 

All  stress  is  given  to  the  flanges  by,  or  through,  the  rivets, 
binding  them  to  the  web,  hence  their  proper  distribution  be- 
comes a  matter  of  importance  ;  it  will  be  shown  by  two  ex- 
amples. 

The  exact  analytical  determination  of  the  web  thickness 
cannot  be  reached,  but  the  following  approximate  analysis  is 
frequently  used. 

It  is  shown  by  the  theory  of  elasticity  that  if  two  planes 
at  right  angles  to  each  other  and  to  a  plane  normal  to  the 
neutral  surface  of  a  bent  beam,  be  so  taken  that  their  intersec- 
tion shall  be  found  in  that  neutral  surface  while  their  common 
inclinations  to  it  is45°,  then  there  will  exist  at  the  neutral  axis 
the  same  intensity  of  stress  on  the  two  planes,  but  one  stress 
will  be  tension  and  the  other  compression.  It  is  farther  shown 
that  the  common  intensity  of  the  two  stresses  is  the  same  as 
that  of  either  the  transverse  or  longitudinal  shear  at  the  same 
point,  which  is  also  known  to  be  |  the  mean  for  the  whole 
section  in  the  case  of  a  solid  rectangular  beam. 

Now,  since  the  intensity  of  shear  at  the  neutral  surface  of 
such  a  beam  is  a  maximum  and  zero  at  the  top  and  bottom 
surface,  and  since  it  has  been  assumed  that  the  entire  web 
takes  the  shear  only,  it  follows  that  if  the  shear  be  assumed 
to  be  uniformly  distributed  tliroughout  any  transverse  section 


FLOOR-BEAMS  AND   STRINGERS.  3^9 

of  the  web,  that  the  latter  may  be  supposed  to  be  composed 
of  an  indefinitely  great  number  of  columns,  each  of  which  is 
an  indefinitely  thin  strip  of  the  web,  making  an  angle  of  45° 
with  the  axis  of  the  beam.  In  a  direction  normal  to  these 
columns  an  equal  intensity  of  tension  will,  of  course,  exist. 

One  of  the  preceding  assumptions  is  an  error  on  the  side 
of  danger,  by  making  the  shear  at  the  neutral  surface  only 
two-thirds  of  its  actual  value  ;  while  the  other,  by  making 
the  shear  at  the  top  and  bottom  surfaces  equal  to  the  mean, 
instead  of  zero,  is  an  error  on  the  side  of  safety,  and  its  in- 
fluence largely  predominates  over  the  former. 

The  elementary  columns  of  the  web  may  be  assumed  to 
have  their  ends  fixed  at,  or  by,  the  flange  rivets  of  a  built 
beam,  and  if  d'  is  the  vertical  depth,  between  rivet  hole 
centres  of  the  two  flanges,  the  length  of  the  elementary 
columns  will  be  : 

1=  d' sec  Of^"  =  \./\\\  d' (i). 

If  5  is  the  greatest  total  shear  at  any  transverse  section, 
A  the  area  of  that  section  of  the  web  ;  then  taking  the  depth 
as  d\  and  s  the  mean  shear,  or : 

5 

these  elementary  columns  will  be  subjected  to  an  intensity 
of  compression  equal  to  s.  Hence  if  t,  the  thickness  of  the 
web,  is  sufficiently  great,  there  may  be  taken  by  Gordon's 
formula : 

^  =  -^^ (^)- 

Or,       '  =  '♦''57731) (4). 

If,  for  wrought  iron,  a  —  3000  and  /=  8000,  there  will  re- 
sult :  

/  =  0.0183 /i/ i (4). 

8000  —  s 


330  DETAILS   OF  CONSTRUCTION. 

The  empirical  constants  for  steel  are  yet  to  be  determined. 

In  applying  Eq.  (4)  to  wrought  iron  plate  girders,  it  will 
be  found  that  the  resulting  values  of  t  are  excessive  for  large 
beams.  The  approximations  already  indicated,  and  the  ad- 
ditional fact  that  the  elementary  columns  are  held  in  place 
throughout  their  whole  length  by  the  tension  in  the  web, 
equal  in  intensity  to  the  compression,  and  at  right  angles  to 
the  latter,  are  suflficient  to  justify  the  anticipation  of  such 
results,  Eq.  (3),  therefore,  has  its  chief  value  as  the  basis  of 
an  empirical  formula  for  the  web  thickness. 

Although  the  web  resists  "  shear,"  it  is  evident,  from  the 
preceding  analysis,  that  the  method  of  failure  of  a  web  will 
be  that  of  buckling,  in  which  the  corrugations  will  be  at  right 
angles  to  the  elementary  columns.  Hence,  if  the  web  is  so 
held  that  these  corrugations  are  prevented,  its  resistance  will 
be  very  materially  increased.  This  is  accomplished  by  rivet- 
ing angles,  usually  in  pairs,  on  each  side  of  the  web  at  proper 
intervals,  so  that  the  web  plate  is  securely  held  between 
them.  The  office  of  these  "  stiffeners,"  it  is  to  be  remem- 
bered, is  simply  to  stiffen  the  web,  and  prevent  its  buckling. 
If  they  are  assumed  to  act  as  struts,  the  transverse  shearing 
strain  in  the  web  at  the  section  considered,  must  in  so  much 
exceed  the  compressive  strain  in  the  stiffener-stiaits,  that  the 
rivets  can  transfer  to  it  its  proper  load,  at  the  same  time  pre- 
supposing a  perfect  condition  of  riveting.  In  reality,  neither 
of  those  conditions  can  possibly  exist. 

These  stiffeners  are  ordinarily  riveted  to  the  web  at  right 
angles  to  the  axis  of  the  beam.  If  no  elementary  column  is 
to  be  without  the  support  of,  at  least,  one  of  these  stiffeners 
in  some  portion  of  its  length,  they  must  be  placed  at  a  dis- 
tance apart,  measured  along  the  axis  of  the  beam  not  greater 
than  the  vertical  depth  between  rivet  hole  centres  ;  and  that 
limit  is  very  commonly  given  in  specifications,  although  it  is 
sometimes  placed  at  once  and  a  half  that  depth. 

About  the  same  amount  of  stiffening  would  be  secured  by 
placing  the  stiffeners  at  45°  with  the  axis  of  the  beam,  and 
at  intervals  of  twice  the  depths,  but  the  difficulties  of  con- 
struction would  be  increased. 


FLOOR-BEAMS  AND   STRINGERS.  331 

A  safe  rule,  and  one  frequently  used,  though  purely  con- 
ventional, is  to  introduce  stiffeners  when  the  mean  shear,  i.  e., 
S  -^A,  exceeds  4000  pounds  per  square  inch  for  wrought  iron. 

The  function  of  the  rivets  holding  the  flanges  to  the  web  is 
next  to  be  considered.  In  reality,  these  rivets  may  have  two 
offtces  to  perform.  If  the  load  of  the  girder  rests  on  one  of 
its  flanges,  the  flange  rivets  will  sustain  it  directly ;  but 
their  chief  office  is  to  give  the  flanges  their  proper  stress.  If 
then  the  stresses  be  determined  for  any  two  points  between 
the  end  of  a  girder,  and  the  point  of  greatest  flange  stress, 
the  shearing  or  bearing  resistance  of  all  the  rivets  between 
those  points  must'be  equal  to,  or  not  less  than  the  resultant  of 
the  difference  between  the  determined  flange  stresses  and 
the  load  resting  on  the  flange  between  the  same  points.  If 
no  load  rests  on  the  flange,  but  is  carried  directly  by  the  web, 
the  "  resultant  "  is  evidently  the  simple  difference  between 
the  determined  flange  stresses. 

These  elementary  considerations  constitute  the  entire 
method  of  finding  the  pitch  and  number  of  rivets  in  the 
flanges  of  built  beams,  and  will  be  applied  to  two  examples. 

The  complete  design  of  the  truss,  treated  in  Art.  ii,  is  to 
be  given  in  subsequent  pages,  and  in  the  present  connection 
the  stringers  and  floor-beams  will  be  discussed.  The  strin- 
gers will  be  placed  7  feet  apart  centres,  and  the  rails  will  be 
laid  on  8  inch  by  8  inch  ties  9  feet  long,  spaced  16  inches 
from  centre  to  centre.  The  ties,  rails,  guard  rails,  splices, 
spikes,  etc.,  will  then  weigh  about  325  pounds  per  lin.  ft. 
The  depth  of  stringers  will  be  taken  at  27  inches  throughout 
their  lengths,  and  the  iron  of  each  stringer  will  be  assumed 
to  weigh  100  pounds  per  lineal  foot.  The  total  fixed  load 
will  then  amount  to  263  pounds  per  lineal  foot  of  each  strin- 
ger, while  the  moving  load  is  one-half  of  the  concentrations 
given  in  the  engine  diagram  of  Art.  ii.  The  principles  es- 
tablished in  Art.  7  show  that  the  four  10,000  pound  driving 
wheel  loads  will  produce  the  greatest  bending  moment  when 
either  wheel  ^  or  ^  is  at  the  distance  1.062  feet  from  the 
centre  of  the  panel,  and  will  be  found  under  the  wheel  in 
question. 


332  DETAILS  OF  CONSTRUCTION: 

With   that  position,  the  following  moving    load   bending 

moments  will  exist : 

« 

Maximum 1 19,200  ft.  lbs. 

7^  feet  from  the  end 105,400  "     " 

5      "       "        "      " 85,600  "     " 

2i    "       "        "      " 44,800  "     " 

It  is  evident  that  the  last  three  of  these  values  are  not  the 
greatest  moments  for  those  points,  but  as  the  flanges  are  to  be 
of  uniform  section,  it  is  not  necessary  to  seek  them,  as  might 
easily  be  done  by  the  aid  of  the  principles  of  Art.  7. 

The  vertical  depth  between  the  rivet  hole  centres  of  these 
stringers  will  be  taken  at  2  feet.  The  flange  stresses  at  the 
various  points  will  then  be  : 

.  26%  X  (20.55)-       IIQ,200       ^^  ,,  „„ 

At  centre   .     .     .         ^  „  ^ ^^  +   -^ =  66,540  lbs.    CD. 

8x2  2  •' 

Iff  1  263   X   7.5   X    13.05         105,400        ^  u         -c-n 

7*ft.fromend — ^ '-^ ^-^-+      ^^      =  59,144       •    ^^• 

'  ■  2x2  2  ^ 

,.       .,  ..  263     X     5     X     15.55  85,600  .,         ^rr 

c     "     "       "        — ^ 2 ^-^  +  • — - —  =  47,920   "  .   GH. 

^  2X2  2 

263x2.5x18.05  ^  ^4^800  ^  ^^^^g^  ,.  _  ^^^ 


^3 


2X2 


The  allowed  working  stresses  in  the  flanges  of  the  stringers 
will  be  taken  at  7,000  pounds  per  sq.  in.  of  gross  section  in 
compression  and  8,000  pounds  per  sq.  in.  of  net  section  in 
tension.  The  diameter  of  rivets  in  the  stringers  and  floor 
beams  of  railway  bridges  is  chiefly  a  matter  of  judgment  ;  it 
usually  ranges  three-quarters  to  seven-eighths  of  an  inch.  In 
the  present  instance,  rivets  of  the  latter  diameter,  before 
being  driven,  will  be  taken.  The  metal  punched  out  for  a 
rivet  should  leave  a  hole  not  more  than  one-sixteenth  of  an 
inch  greater  in  diameter  than  that  of  the  cold  rivet.  But 
the  metal  immediately  about  the  edge  of  the  hole  is  materi- 


FLOOR-BEAMS  AND   STRINGERS.  33^ 

ally  injured  for  tensile  purposes,  and  in  the  tension-chord 
angles  the  disc  of  metal  rendered  valueless  should  be  taken 
one-eighth  of  an  inch  greater  in  diameter  than  that  of  the 
cold  rivet,  i.e.,  for  the  present  case,  one  inch.  In  the  com- 
pression flange  no  metal  need  be  deducted  for  the  rivet 
holes. 

The  upper  flange  section  at  the  centre  will  be  : 

66,540  -4-  7,cxx)  =9.5  sq.  in. 

The  net  lower  flange  section  will  be  : 

66,540  ^  8,000  =  '^.l  sq.  in. 

Tlic  following  flanges  will  satisfy  the  conditions  : 

ttj.^      n  i  2  —     5"  X  4''  30  lb.  angles. 

Upper  flans^e     .     .     .     .       ■  „         „  , 

(  I  —  10    X  I    cover  plate. 

Lower  flange    ....         2  —    5"  x  ^'  ^"j  Va.  angles. 

The  62  lb.  angles  have  a  thickness  of  three-quarters  of  an 
inch,  so  that  the  excess  of  gross  section  exactly  covers  the 
metal  destroyed  by  the  punch. 

It  is  the  best  practice  to  run  a  plate  the  whole  length  of 
the  upper  flange  in  order  to  make  it  act  as  a  unit  in  resisting 
compression.  The  lower  flange  needs  no  cover  plate,  and, 
again,  the  additional  rows  of  rivet  holes  would  produce  more 
dead  metal. 

Although  there  is  a  little  waste  of  metal  near  the  ends  in 
small  plate  girders,  it  is  economy  to  save  labor  by  making  the 
flanges  of  uniform  section  throughout  their  lengths.  This 
economy  ceases  when  the  flanges  become  so  heavy  that  one 
or  more  cover  plates  become  necessary  in  the  tension  flange 
and  more  than  one  in  the  compression  flange. 

Cover  plates  should  be  carried  at  least  a  foot  beyond  the 
section  at  which  the  additional  metal  is  required,  and  rivets 
should  be  closely  pitched  in  that  portion.  If  rivets  pierce 
both  legs  of  an  angle  in  tension,  metal  should  be  deducted 
for  all  the  rows  in  both  less. 


334  DETAILS  OF   CONSTRUCTION. 

The  locatiori  of  the  preceding  sections  is  shown  in  Fig.  I, 
of  PI.  XI.  The  increments  of  stresses  for  the  different  seg- 
ments of  the  flanges  will  be  : 

EC  or  FD  —  66,540  —  59,144  =    7'396  lbs. 
GE  or  HE  =  59,144  —  47,920  =  1 1,224   " 
GK  or  HL  =  47,920  —  25,367  =  22,553    " 
KA   or  BL   ^  =  25,367    " 

The  extent  of  flange  over  which  the  weight  of  each  driving 
wheel  weighing  10,000  pounds  will  be  distributed  is  indeter- 
minate, but  as  the  ties  are  16  inches  apart  centres,  it  will  be 
sufTficiently  accurate  to  take  that  distance  as  2i  feet.  As 
each  driving  wheel  passes  over  the  entire  stringer,  the  result- 
ant stresses  which  the  rivets  in  the  various  sections  into  which 
the  beam  is  divided  will  be  obliged  to  carry,  are  as  follows: 

EC  or  ED  =  y'(  7,396)^  +  (10,000)^  =  12,440 

GE  or  HE  =  y'(i  1,224)^  +  (10,000)-  =  15,000. 

GK  or  HL  —  ^(22,553)^  +  (10,000)-  =  24,700. 

KA  or  BL   —  y/{2^,^(^'jj  +  (10,000)^  =  27,300. 

In  order  to  determine  the  number  of  rivets  in  any  of  these 
sections,  it  will  be  necessary  to  fix  the  thickness  of  web  plate. 
The  minimum  thickness  permissible  is,  to  some  extent,  a 
matter  of  judgment,  but  it  is  safe  to  say  that  no  built  beam 
for  railroad  purposes  should  have  a  less  thickness  of  web 
than  YF  of  an  inch,  and  a  limit  of  |  is  still  better  practice. 
The  latter  limit  will  be  used  here. 

With  the  position  of  moving  load  already  determined  in 
Art.  II,  the  greatest  shear  at  the  end  of  the  stringer  is 
24,830  +  2,700  =  27,530  pounds  =  S.  The  sectional  area 
/i  of  a  27  by  I  inch  plate  is  10.125  sq.  ins.  =  A.  Hence  s  — 
S  -^  A  —  2,720  pounds  ;  also  /  =  24  x  1.414.  =  33-94  inches. 
Hence  Eq.  (4)  gives: 

/  =  0.45  inch. 

Since  this  result  is  greater  than  the  assumed  value  of  /,  the 


FLOOR-BEAMS  AND    STRINGERS.  335 

hypothetical  elementary  columns  are  not  capable  of  sustain- 
ing their  loads  without  exceeding  by  a  little  the  proper  work- 
ing stress ;  but  as  the  hypothesis  involves  a  considerable 
safety  error,  the  assumed  value  of  t  is  probably  ample.  How- 
ever, as  the  additional  metal  is  very  small  in  amount,  a  pair 
of  3  by  2\  1 6-pound  L  stiffeners  will  be  introduced  at  the 
distance  of  27  inches  from  each  end  as  shown. 

The  working  resistance  to  shearing  offered  by  rivets  in  the 
truss  will  be  taken  at  7,500  pounds  per  sq.  in.,  and  the  limit- 
ing pressure  between  rivets  and  walls  of  holes  will  be  fixed  at 
12,000  pounds  per  sq.  in.  under  the  same  circumstances.  The 
floor  of  a  bridge  is  subject  to  shocks  due  to  track  imperfec- 
tions, and  the  above  values  should  be  reduced  by  25  per  cent., 
making  the  working  resistance  to  shearing  5,625  pounds,  and  to 
pressure  9,000  pounds  per  sq.  in.  As  the  web  is  embraced  by  an 
angle  iron  on  each  side,  each  rivet  will  be  subjected  to  double 
shear,  and  the  thickness  of  bearing  surface  of  the  web  will 
be  much  less  than  that  of  the  two  flange  angles.  In  the  de- 
termination  of  the  shearing  and  bearing  resistances  of  rivets, 
the  cold  diameter,  before  being  driven,  should,  as  a  margin 
of  safety,  be  considered.  Hence,  those  resistances  for  the 
seven-eighths  rivets  under  treatment,  are: 

2.\.  71  (0.875)^  •  5,625  =  6,750  pounds;  and, 
0.875  ■  0.375  •  9'000  =  3,000      " 

The  latter  quantity  is  much  the  smaller,  and  will  govern 
the  number  of  rivets  in  each  section,  as  follows : 

EC  or  FD 12,440  ^  3,000  =  4  rivets  required. 

6^£  or ///^.  ...  15,000 -^  3,000  =  5       "  " 

GKox  HL 24,700 -^  3,000  =  8      "  " 

KA  or  BL 27,300  ^  3,000  =  9      "  " 

The  nearest  whole  number  is  taken  in  each  case. 

These  results  give  at  and  near  the  end  about  nine  rivets  to 
each  two  and  a  half  feet.  A  uniform  pitch  of  three  inches, 
therefore,  will  be  assumed  throughout   each   flange.     If  the 


33^  DETAILS  OF  CONSTRUCTION. 

load  is  uniform  in  intensity,  it  is  well  known  that  the  varia- 
tion of  flange  stress  is  very  little  for  a  considerable  distance 
either  side  of  the  centre.  This  example  shows,  however,  that 
concentrated  loads  may  require  just  as  close  centre  riveting, 
i.e.,  just  as  small  pitch,  as  at  the  ends.  The  number  of  sec- 
tions into  which  a  beam  must  be  divided  is  a  matter  of  judg- 
ment in  each  particular  case. 

The  greatest  shear  at  the  end  of  the  stringer  has  already 
been  seen  to  be  27,530  pounds,  hence  the  end  stifTeners  at  A 
must  transfer  that  amount  to  the  floor-beam  web.  An  inten- 
sity of  4,000  pounds  per  square  inch  of  normal  section  is  a  safe 
and  proper  value  for  those  members.  The  end  stiffeners  will 
then  be  assumed  to  be  2  —  4"  x  4"  35  lb.  angles,  one  being 
on  each  side  of  the  web  at  each  end  of  the  beam  and  extend- 
ing the  full  depth  between  the  legs  of  the  flange  angles  of  the 
latter.  Fillers  whose  thickness  is  just  a  little  less  than  that 
of  the  heaviest  flange  angle  will  be  required  under  these  end 
stiffeners.  Light  intermediate  stiffeners  may  be  bent  to  fit 
if  the  flange  angles  are  not  too  heavy ;  otherwise  fillers  must 
again  be  used. 

The  number  of  rivets  required  to  transfer  the  greatest  re- 
action at  the  stringer  ends  to  the  floor-beam  (see  Fig.  2,  PI. 
XT.)  is  found  by  taking  the  reaction  of  the  locomotive  load 
(found  in  Art.  11  to  be  34,600  pounds)  and  adding  to  it  the 
fixed  load,  or,  20.55  ^  263  =  5,405  pounds,  then  dividing  the 
result  by  the  bearing  capacity  of  seven-eighths  rivet  in  the 
three-eighths  web  of  the  floor-beam,  as  that  is  less  than  the 
resistance  of  the  same  rivet  in  double  shear.  The  number  of 
rivets  needed  will  then  be  (34,600  +  5,405)  -^  3,000  =  13. 
Seven  rivets  will  be  placed  in  each  4x4  end  angle,  as 
shown  in  Fig.  2,  of  PI.  XL,  making  14  for  the  end  of  each 
stifl"ener.  If  the  loads  were  very  great,  it  would  be  neces- 
sary to  reinforce  the  web  plate  of  the  floor-beam  where 
it  receives  the  ends  of  the  stiffeners,  by  riveting  a  plate 
on  each  side.  In  the  present  instance,  however,  it  is  un- 
necessary. 

The  bill  of  material,  with  the  weights  of  one  stringer,  may 
now  be  written  as  follows  : 


FLOOR-BEAMS  AND    S7AVNGERS. 


337 


1  27  X    I  Plate  20.55  ft.  long 695  lbs. 

2  5x4  30  lb.  angles     "      "      "    411 

2     5  X    4  47  "         "  "      "      "    644 

I    10  X   I  Plate  "      "      "    257 

4     4  X  1^  Filling  plates  19  ins.      "    60 

43x4        •'  "       19   "  " 32 

4     4  X    4  35  lb.  angles  27  "  "    105 

4     3  X  2|  16  "       "        27  "  "    48 

5  Rivets 1 20 


Total  weight  of  stringer 2,372  lbs. 

The  actual  weight  per  foot  is  thus  about  15  lbs.  more  than 
was  assumed.  Although  this  makes  an  insignificant  differ- 
ence in  the  flange  section,  and  is  amply  provided  for  in  the 
present  instance,  in  practice  the  actual  weight  should  be  a 
little  under  the  assumed,  and  not  over  it. 

The  arrangement  of  connecting  the  stringers  to  the  floor- 
beams  shown  in  Fig.  2  of  PI.  XI.,  has  the  merit  of  making  a 
very  stiff  floor  system.  It  is  proper  to  say,  however,  that 
many  other  methods  are  used.  The  small  angle  brackets 
seen  under  the  ends  of  the  stringers  and  riveted  to  the  lower 
flange  angles  of  the  floor-beam  are  simply  for  convenience  in 
erection,  and  are  not  considered  as  essential  to  the  resistance 
of  the  joint. 

According  to  the  preceding  bill  of  material,  the  actual 
maximum  weight  concentrated  at  the  stringer  ends  will 
be: 

34,6cx)  +  20.55   ^  300  =  40,800  nearly. 


The  depth  of  the  floor-beam  will  be  taken  at  36  inches 
throughout  its  entire  length,  so  that  the  vertical  distance  be- 
tween rivet  centres  in  the  two  flanges  will  be  about  33  inches. 

If  the  weight  of  the  floor-beam  be  assumed  at  150  lbs.  per 
lineal  foot,  the  total  flange  stresses  at  the  centre,  at  the 
stringer  points,  and  at  points  2^  feet  distant  from  the  ends, 
will  be  : 


338  DETAILS  OF  CONSTRUCTION. 

40,800  X  5        150  X  (17)^  _ 

2.75  8  X  2.75    ~  ^  '^  ' 

40,800  X   5  iSOJiA        ^j3      ^5840, 

2.75  2   X    2.75  /:?'   t   » 

40,800  X  2. 5       150  X  2.5  _        _ 

^t_! 2  +   ^ ±      X   14.5  =  38,090. 

2.75  2    X   2.75  ^^        -^    '^ 

Using  the  same  working  stresses  in  the  flanges  as  were 
fixed  for  the  stringers,  there  will  be  found  for  the  upper 
flange  section  at  the  centre : 

76,170  -f-  7,000  =  10.9  sq.  ins. ; 

and  for  the  net  lower  flange  section  : 

76,170  ^  8,000  —  9.5  sq.  ins. 

The  following  flanges  will  satisfy  these  requirements : 

(  2  —     5"  X  4"  36  lb.  angles, 

^^     ■'  1  I  —  12     X  tV  cover  plate. 

Lower  flange 2  —     5"  x  4"  55  lb.  angles. 

The  thickness  of  the  55  lb.  angle  is  about  0.7  inch,  so  that 
if  seven-eighth  cold  rivets  are  used  the  metal  destroyed  by 
the  punch  is  just  about  equal  to  the  excess  of  the  total  section 
over  the  net. 

The  total  shear  at  the  end  of  the  floor-beam  is  40,800  +  8.5 
X  150  =  42,075  lbs.  =  5.  If  the  web  thickness  be  assumed 
at  three-eighths  of  an  inch,  /^  =  36  x  |  =  13,5  sq.  ins. 
Hence,  s  =■  S  ^  A  —  42,075  -f-  13.5  =  3,120  lbs. 

/  =  33  X   1,414  —  46.66.     These  quantities  inserted  in  Eq. 

(4)  give : 

t  =  0.68  inch. 

This  value  shows  that  if  the  hypothetical  elementary 
columns  are  to  sustain  3,120  lbs  to  the  sq.  in.,  the  web  must 


FLOOR-BEAMS  AND   STRINGERS.  339 

be  0.68  inch  thick.  But  such  a  thickness  is  plainly  excessive, 
and  shows  how  the  formula  errs  in  the  direction  of  safety. 
A  web  thickness  of  three-eighths  of  an  inch  will  be  assumed, 
and  3"  X  2\"  16  lb.  angle  stifTeners  placed  half-way  between 
the  stringer  supports  and  the  ends,  as  shown  at  EF,  Fig.  2, 
PI.  XI.  These  stiffeners  will  require  3"  x  |"  filling  plates  28 
inches  long. 

If  shearing  and  bearing  resistances  of  5,625  and  9,000  lbs. 
per  sq.  in.,  respectively,  be  taken,  as  was  done  in  designing 
the  stringers,  the  bearing  value  of  one  rivet  in  the  three- 
eighths  web  will  be,  as  before,  0.875  ^  0.375  x  9,000  =  3,000 
pounds. 

Hence  the  number  of  seven-eighths  rivets  required  between 
the  three  sections  of  the  beam  will  be  : 


Centre  and  C (76,170  —  75,840) 

C  and  EF (75-840  —  38,090) 

EF''   AB 38,090 


3,000=  I, 
3,000=  13, 
3,000  =  13. 


These  conditions  will  be  sufificiently  near  fulfilled  if  a  pitch 
of  2|  inches  be  taken  for  a  distance  of  6  feet  from  each  end, 
and  six  inches  for  the  remaining  5  feet  between  the  stringers. 
The  flange  stresses  do  not  require  a  six-inch  pitch  at  the 
centre,  but  that  value  should  not  be  exceeded,  in  order  that 
the  flanges  may  be  properly  bonded.  The  pitch  in  the  cover 
plate  on  the  upper  flange  should  be  2f  inches  for  a  distance 
of  15  inches  from  each  end,  and  over  the  remaining  portion 
of  its  length  that  pitch  may  be  doubled.  In  no  case,  how- 
ever, should  the  pitch  in  a  compression  plate  exceed  about  16 
times  its  thickness. 

The  end  floor-beams  are  suspended  from  the  pins  by  the 
plates  shown  riveted  to  the  heavy  end  stifTeners,  as  at  AB. 
Those  StifTeners  transfer  half  the  weight  of  the  beam  in  addi- 
tion to  the  40,800  pounds  at  the  stringer  ends,  to  the  suspen- 
sion plates,  or,  40,800  4-  1 50  x  8.5  =  42,100  lbs.  As  these 
end  stiffeners  do  not  in  this  instance  sustain  this  entire  weight 
at  any  section,  the  latter  cannot  be  analytically  determined. 
They  should  be  heavy,  however,  and  will  be  taken  as  5  x  4 
.36  lb.  angles. 


340  DETAILS  OF   CONSTRUCTION. 

The  number  of  rivets  required  between  the  end  stiffeners 
and  the  web  is  42,100  h-  3,000  =  14,  but  it  is  convenient  to 
take  15  as  shown. 

The  dimensions  of  the  suspension  plates  cannot  be  fixed 
until  the  diameter  of  the  pin  is  determined,  and  they  will  be 
found  in  a  later  Art. 

The  bill  of  material  for  the  floor-beam  with  its  weight  will 
now  be  : 

1  36  X  I  Plate  17  ft  long  17  x  45     —  765 

2  5  X  4  36  lb.  angles     "    "     "  34  x   12     =  408 

1  12  X  ^  cover  plate     "    "     "  17  x   12.5=  213 

2  5  x  4  55  lb.  angles  "  "  "  34  x  i8|^  =  623 
4  3  x  f  filling  plates  2\  "  "  9^  x  ^\  =■  58 
4  3  X  2|  16  lb.  stiffening  angles  3  ft.  long  4  x  16  =  64 
4  5  X  I  end  "  plates  2i"  "  10.4  x  81  =  87 
4  5x4  36  lb.  angles  3  "  "  4  x  36  =  144 
4  3  X  2^  16   "      "                        I    "     "      i^    X  16    =     21 

\  rivets =130 


Total  weight  of  beam     ....     2,513  lbs. 

The  weight  per  lineal  foot  is  then  2,513  -^  17  =  148 
pounds,  or  less  than  that  assumed,  as  it  should  be. 

In  the  cases  of  large  plate  girders  it  is  necessary  to  make 
splices  in  the  web  plate,  a  splice  plate  being  used  on  each 
side  of  the  web.  The  combined  thickness  of  these  splice 
plates  should  be  at  least  50  per  cent,  in  excess  of  that  of  the 
plates  spliced.  The  number  of  rivets  on  each  side  of  the  joint 
should  be  such  that  the  total  bearing  resistance  in  the  web,  or 
the  total  shearing  resistance  of  the  rivets,  shall  at  least  equal 
the  greatest  possible  transverse  shear  at  the  joint  considered. 
At  least  two  rows  of  rivets  should  be  found  on  either  side  of 
the  joint. 

It  is  sometimes  customary,  if  web  plates  have  no  splices,  to 
take  one-sixth  of  the  web  section  as  acting  in  either  flange. 
If  no  rivet  holes  were  punched  for  the  stiffeners,  this  method 
would  be  allowable.    But  such  rivet  holes  frequently  take  out 


FLOOR-BEAMS  AND    STRINGERS.  341 

considerable  metal,  and  as  the  tension  side  of  the  plate  only 
is  affected,  one-sixth  of  the  remaining  metal  ceases  to  be  a 
proper  proportion.  On  the  whole,  therefore,  it  is  better  to 
neglect  the  bending  resistance  of  the  web,  and  allow  it  to 
balance,  so  far  as  it  may,  the  effect  of  the  rivet  holes  being 
out  of  the  centre  of  gravity  of  the  flange  angles. 

That  depth  of  plate  girder  Avhich  will  give  the  least  weight 
depends  entirely  upon  the  manner  and  amount  of  the  load- 
ing. With  very  heavy  concentrated  loads,  it  may  be  half 
the  span ;  on  the  other  hand,  with  very  light  loads  it  may  be 
less  than  one-twentieth  of  the  span.  As  all  bending  moments 
may  be  supposed  to  be  caused  by  some  uniform  load,  either 
fictitious  or  real,  the  following  analytical  discussion  may  be 
of  some  value  as  well  as  interest : 

Economic  Depth  of  Plate  Girders  with  Uniform  Flanges. 

If  a  plate  girder  carries  a  uniform  load,  and  is  designed 
with  flanges  of  uniform  cross  sectional  area,  the  depth  which 
will  give  the  least  weight  of  girder  may  easily  be  obtained. 

Let    /  =  span  in  feet. 

d  —  depth  "     '• 

t  —  web  thickness  in  inches. 

/  =  allowed  working  stress  in  lbs.  per  sq.  in.  for  the  flanges. 
/'  =        "  "  "         ''  "        end  stiffeners. 

a  =  sectional  area  of  each  intermediate  stiffener  in  jr^.  _/y. 

n  =  number  of  stiffeners  (intermediate). 
lu  =  total  load  per  lin.  ft.  of  girder  (in  pounds). 

The  flange  stress  at  centre  will  then  be : 

F=  — 
%d' 

The  volume  of  the  web  plate  in  cu.  ft.  will  be : 

12* 


342  DETAILS   OF   CONSTRUCTION. 

If  one-sixth  of  the  latter  is  taken  to  be  concentrated  in  the 
flange,  the  volume  of  the  two  flanges  in  cu.  ft.  will  be: 

2FI  I  Idt  _    wP         Idt 

144/  6  \2  ~  S7^P^       36' 

The  volume  of  the  end  stiffeners  will  be : 

wld 
144/' 

.and  that  of  the  intermediate  stiffeners: 

nad. 

The  volume  of  the  entire  girder  will  then  take  the  value: 

-,        wP         Idt        wld  ,  ,  , 

V— — -^—j  +  --  +  ,  +  nad .     .......     (5). 

576/(3^       18        144/  ^^' 

By  taking  the  first  derivative ; 

dV  _  _      wP  It  wl  _ 

did)  ~        S7^pd^  ^  18  ^  144/^  +  na-0. 

Solving  for  d^: 


^    ,  //  wl 

S7^P  {  —o   +   Ti  +  ^^^ 

^  m8       144/ 


.d='-A/ ^ .     .     .     (6). 

^    /(8//  +  ^  +  I44««) 

If  one-sixth  of  the  web  is  not  concentrated  in  each  flange, 
12//  will  take  the  place  of  8//  in  Eq.  (6). 

If  all  stiffeners,  both  end  and  intermediate,  are  omitted, 
Eq.  (6)  will  take  form  : 

.=  ^Y|.     .....     (7). 

In  reality  /  is  seldom  or  never  exactly  the  same  for  both 


EYE-BARS   OR   LINKS.  343 

flanges,  since  it  is  the  working  stress  in  reference  to  the  gross 
section.  It  will  be  sufficiently  near,  however,  for  all  usual 
purposes  to  make  it  a  mean  of  the  two  actual  working  stresses 
for  the  gross  sections. 

It  should  be  borne  in  mind  that  local  circumstances  fre- 
quently compel  a  different  depth  from  that  given  by  Eq.  (6). 
It  will  also  be  found  that  a  considerable  variation  from  that 
depth  will  cause  a  comparatively  small  variation  in  weight. 
Again,  the  difficulties  of  handling  a  deep  girder,  and  the  shop 
cost  per  pound,  may,  and  usually  does,  make  the  economic 
depth  a  little  less  than  that  found  by  the  aid  of  Eq.  (6). 

Art.  73. — Eye-Bars  or  Links. 

An  "eye-bar"  or  "link"  is  a  tension  member  of  a  pin- 
connection  bridge,  fitted  at  each  end  with  an  eye  for  the  in- 
sertion of  a  pin.  Two  views  of  an  eye-bar  are  shown  in  Fig. 
5,  PI.  XII.;  A  is  the  body  of  the  bar,  D  the  neck,  and  C  the 
eye.  The  head  of  an  eye-bar  is  the  enlarged  portion  in  which 
the  pin-hole  is  made.  The  eye-bar  is  one  of  the  most  im- 
portant members  of  a  pin-connection  bridge,  and  the  deter- 
mination of  the  relative  dimensions  of  the  head  has  been  the 
subject  of  much  experimenting.  A  mathematical  investiga- 
tion; however,  with  the  same  object  in  view  is  a  matter  of 
considerable  complexity,  although  an  approximate  solution 
of  the  problem  may  be  obtained,  and  its  agreement  with  the 
results  of  experiment  is  quite  close. 

Before  taking  a  general  view  of  the  stresses  which  may 
arise  in  an  eye-bar  head,  it  must  be  premised  that  a  difference 
of  ^V"  to  ^V"  between  the  diameter  of  the  pin  and  that  of 
the  pin-hole  is  considered  exceedingly  good  practice.  Before 
the  eye-bar  is  strained,  therefore,  there  is  a  line  of  contact 
only  between  the  pin  and  eye-bar  head,  but  on  account  of 
the  elasticity  of  the  material,  this  line  changes  to  a  surface 
when  the  bar  is  under  stress,  and  increases  with  the  degree  of 
stress  to  which  the  bar  is  subjected.  This  line  and  surface  of 
contact  is,  of  course,  in  the  vicinity  of  K,  Fig.  3,  PI.  XII.,  i.e., 
on  that  side  of  the  pin  toward  the  nearest  end  of  the  bar. 
The  consequence  of  this  is  that,  when  the  bar  is  strained,  the 


344  DETAILS   OF  CONSTRUCTION. 

portion  about  KA,  Fig.  3,  is  subjected  to  direct  compressiort 
and  extension ;  that  about  BL,  DH,  and  FM  to  direct  ten- 
sion and  bending,  while  in  the  vicinity  of  CN  (also  CQ)  there 
is  a  point  of  contra-flexure,  and  the  stress  in  the  direction  of 
the  circumference  changes  from  compression  to  tension  as  H 
is  approached  from  K. 

It  should  have  been  said  before  that  if  w  represents  the 
width  of  an  eye-bar,  as  shown,  then  its  thickness,  /,  is  gener- 
ally included  between  the  limits  \w  and  \w.  These  limits 
of  the  relative  values  of  the  quantities  are  seldom  exceeded. 

Fig.  2,  PI.  XII.,  represents  a  method  of  laying  down  an 
eye-bar  head  which  has  been  determined  by  a  very  extensive 
system  of  experiments  given  by  a  member  of  the  British  In- 
stitution of  Civil  Engineers,  and  one  that  has  stood  the  test 
of  long  American  experience  ;  in  short,  there  is  probably  no 
better  method  known.  Let  r  represent  the  radius  of  the  pin- 
hole, and  w  the  width  of  the  bar. 

Then  take  EN  =  o.66iu.  The  curve  DRBK  is  a  semicircle 
with  a  radius  equal  to  r  4-  0.66w,  with  a  centre,  A,  so  taken 
on  the  centre  line  of  the  bar  that  QB  =  0.87a'.  GF  is  a  por- 
tion of  the  same  curve,  with  A'  as  the  centre  (yi'C=  y^Q  ; 
GH'xs  any  curve  with  a  long  radius  joining  6^/^  gradually  with 
the  body  of  the  bar.  HG  should  be  very  gradual  in  order 
that  there  may  be  a  large  amount  of  metal  in  the  vicinity  of 
CG,  for  there  the  metal  is  subjected  to  flexure  as  well  as 
direct  tension.  FD  is  a  straight  line  parallel  to  the  centre 
line  of  the  bar. 

Fig.  3  shows  another  method  founded  on  the  results  of  a 
mathematical  investigation.  Take  r  and  w  as  before.  Then 
BC=  AC  =  r  +  0.S77V,  DH—  \w  =■  o.66w,  ED  =  EF=  2r  + 
w.  DF  is  described  with  ED  until  DCF=  45°.  BAB  is  de- 
scribed with  BC  until  BCA  =  35°.  BN  is  drawn  from  Z  as  a 
centre  located  in  such  a  position  as  to  cause  that  arc  to  be  at 
the  same  time  tangent  to  j97Vand  AB.  DN  \s  a  straight  line 
drawn  parallel  to  the  axis  of  the  bar.  PF'xs  any  easy  curve 
which  will  appear  the  best.  The  dotted  lines  in  both  Fig.  2 
and  Fig.  3  show  the  slope  that  should  be  given  in  order  to 
clear  a  die. 


SIZE  OF  PINS.  345 

The  outline  of  the  head  is  now  usually  formed  of  a  portion 
of  the  circumference  of  a  circle  whose  centre  is  the  centre  of 
the  pin-hole.  In  such  a  case  no  dimension  of  the  head  should 
be  less  than  the  corresponding  one  determined  by  either  of 
the  methods  just  given. 

Fig.  4  shows  the  head  thickened  in  such  a  manner  that  the 
mean  maximum  intensity  of  pressure  between  pin  and  pin- 
hole shall  not  exceed  a  given  amount,  p.  Let  7"  represent 
the  maximum  intensity  of  tension  in  the  body  of  the  bar; 
then,  as  has  been  shown  in  discussing  the  pressure  against 
the  bodies  of  rivets  : 

wtT 


wtT=  2rpt'  .-.  t' 


2rp 


Art.  74.— Size   of  Pins. 

The  exact  analytical  determination  of  the  pin  diameter  in 
any  particular  case  is,  like  many  other  matters,  involving  the 
elasticity  of  materials,  an  impossibility,  although  the  problem 
in  its  simplest  form  was  subject  to  a  very  able  mathematical 
investigation  by  Charles  Bender,  C.  E.,  in  Van  Nostrand's 
Magazine  for  October,  1873.  One  or  two  reasonable  assump- 
tions, which,  in  a  great  majority  of  cases,  must  be  very  nearly 
accurate,  give  the  problem  a  very  simple  character.  The  first 
of  these  assumptions  is  that  tJic  pressure  applied  to  any  pin  has 
its  centre  at  the  eentre  of  the  surface  of  contact.  Fig  3  of  PL 
XI.  represents  the  lialf  of  a  pin-connected  joint,  LL  being  the 
centre  line,  and  by  this  assumption  the  centre  of  pressure  be- 
tween each  of  the  eye-bars  A,  B,  C,  D,  etc.,  and  post  bearing 
P,  and  the  pin  is  located  half-way  between  the  faces  of  those 
members  normal  to  the  axis  of  the  pin. 

If,  however,  a  pin  is  held  by  a  compression  member,  such 
as  an  upper  chord  or  post,  then  the  centre  of  pressure  in  that 
member  may  be  taken  as  the  eentre  of  such  a  surface  as  zvill 
reduce  the  bearing  intensity  to  its  maximum  limit. 

It  is  to  be  premised  that  the  general  considerations  touch- 
ing the  distribution  of  pressure  between  rivets  and  plates 
given  in  Art.  70  hold  equally  true  for  pins.     The  greatest  al- 


346  DETAILS  OF  CONSTRUCTION 

lowable  bearing  intensity  between  pins  and  eye-bars  of 
wrought  iron  ranges  from  io,ooo  to  12,500  pounds  per  square 
inch  of  the  surface  found  by  multiplying  the  diameter  of  pin 
by  thickness  of  bar.  The  latter  product  is  always  considered 
the  bearing  surface. 

If  two  bars  only,  such  as  A  and  D,  act  on  each  end  of  a 
pin  it  is  clear  that  the  centre  line  of  the  latter  will  be  convex 
toward  D.  The  result  will  be  a  movement  of  the  centres  of 
pressure  of  those  bare  toward  each  other ;  so  that  the  lower 
arm  of  A  will  be  less  than  half  the  thickness  of  that  member 
plus  half  that  of  B.  The  second  assumption  given  above 
seems  thus  very  reasonable,  and  may  be  extended  to  the  case 
of  a  pair  of  eye-bars,  only,  at  the  end  of  a  pin.  When,  on 
the  other  hand,  a  number  of  eye-bars  of  various  sizes  take 
hold  of  a  pin,  particularly  if  the  bending  moments  have  dif- 
ferent directions  at  different  sections  of  the  pin,  the  axis  of 
the  latter  may  be  essentially  straight  and  the  centres  of  press- 
ure should  be  taken  according  to  the  first  assumption.  This 
is,  in  reality,  the  best  practice  in  all  cases,  for  if  the  centre  of 
pressure  departs  from  the  axis  of  the  bar,  the  latter  will  be 
subjected  to  a  bending  moment  equal  to  the  tension  in  the 
bar  multiplied  by  the  distance  of  the  centre  of  pressure  from 
its  axis.  Hence  the  necessity  of  so  fixing  the  diameter  of 
pin  that  it  shall  be  as  stiff  as  possible. 

In  Fig.  3  of  PI.  XL,  let  a  be  the  distance  between  the 
centres  of  eye-bars  A  and  D  ;  a,  that  between  D  and  B  ;  a'' 
that  between  B  and  £,  etc.,  etc.  These  distances  a,  a',  a", 
etc.,  should  always  be  taken  as  the  thickness  of  the  head  plus 
one-eighth  of  an  inch ;  the  latter  amount  representing  about 
the  proper  clearance  in  the  best  work. 

Then  let  Ta,  Ta,  T^,  etc.,  represent  the  total  tensions  in  the 
bars  A,  D,  B,  etc.  The  bending  moments  about  the  centres 
of  those  bars  will  then  be  : 

About  centre  of  D ...  .a  T^ 

''   B   ...  {a  +  a')  Ta  -  T^a'. 
"         "       ''   E .  .  .  .  {a  +  a  +  a")  7^  +  T(,a"  —  Ta{a' +  a"). 
Etc.,    etc.,    etc.,  etc.,  etc. 


SIZE  OF  PINS.  347 

The  rod  7?  is  a  counter  and  does  not  usually  act  when  the 
pin  receives  its  greatest  bending. 

The  preceding  moments  are  all  similarly  formed  and  are 
about  vertical  axes  until  the  centre  of  the  post  bearing,  P,  is 
reached.  The  tension  T  of  the  main  tension  brace  T  pro- 
duces a  moment  about  an  axis  normal  to  its  own.  Let  it  be 
supposed  that  the  resultant  moment  of  all  the  chord  members 
A,  B.^  C,  D  and  E  about  the  centre  of  P  is  right-handed  look- 
ing vertically  down,  as  shown  by  M'  in  Fig.  i. 

Let  M'  represent  that  moment  by  any  con- 
venient scale.  The  moment  of  T  by  a!°,  or  7^", 
will  be  right-handed  looking  upward  ;  and  let 
Mt  represent  that  moment  by  the  same  scale 
as  before.  The  latter  line  is  drawn  normal  to 
the  axis  of  the  member  T.  The  line  M  will 
now  represent  by  the  same  scale  the  moment 
to  which  the  pin  is  subjected  at  the  centre  of 
P,  and  its  direction  is  that  of  the  axis  of  the 
moment.  ^^^-  ^' 

The  greatest  pin  bending  in  the  lower  chord  will  usually 
take  place  with  the  greatest  chord  stresses,  but  the  upper 
chord  pins  will  receive  their  greatest  moments  by  the  great- 
est web  stresses. 

When  a  number  of  bars  are  coupled  to  the  pin  in  such  a 
joint  as  that  shown  in  Fig.  3  of  PI.  XL,  it  is  usually  necessary 
to  test  a  number  of  sections  in  order  to  find  the  greatest  mo- 
ment; unless  the  bars  are  very  nearly  of  the  same  size  and 
placed  alternately  as  shown  when  the  greatest  moment  will 
be  found  at  the  centre  of  the  pin. 

It  is  frequently  advisable,  however,  to  employ  different 
sized  bars  in  order  to  reduce  the  bending  moments  ;  a  small 
bar  being  placed  at  the  end  of  the  pin. 

The  same  reduction  of  bending  moments  is  brought  about 
even  more  effectually  by  the  arrangement  of  lower  chord  bars 
shown  in  Fig.  2. 

In  that  figure  it  will  be  observed  tJiat  the  lower  chord  eye- 
bars  are  so  grouped  on  any  one  pin,  that  the  stresses  in  them- 
for    eaeh   half  of  the  pin,  form  couples  which  have  opposite 


348 


DETAILS   OF  CONSTRUCT  ION. 


signs  and,  thus,  to  a  great  extent,  or  wholly,  neutralize  each 
other. 

By  varying  the  sizes  or  thicknesses  of  the  bars  and  by  re- 
sorting to  the  method  of  grouping  shown  in  Fig.  2  (which 


Fig.  2. 

represents  a  portion  of  an  actual  lower  chord)  the  bending 
moments  in  lower  chord  pins  may  easily  be  reduced  to  any 
desired  extent  in  any  case  whatever. 

It  is  evident  that  the  resultant  moment  shown  in  Fig.  i 
could  be  obtained  by  resolving  the  stress  T  into  its  vertical 
and  horizontal  components  and  combining  their  moments  with 
those  of  the  lower  chord  stresses,  making  the  components 
of  M  vertical  and  horizontal  instead  of  vertical  and  inclined. 

The  bending  of  pins  is  very  much  increased  by  thickened 
eye-bar  heads,  since  the  thickening  increases  the  lever  arm  of 
the  tensile  stress  in  the  eye-bar.  The  thickened  eye  is  a  most 
excellent  thing  for  the  bar,  but  necessitates  an  increased 
diameter  of  pin. 

The  preceding  operations  illustrate  the  general  method  of 
finding  the  bending  moment  to  which  a  pin  is  subjected  in  all 
cases ;  the  component  moments  are  determined  from  the 
stresses  in  the  individual  truss  members,  and  the  resultant  is 
then  found  by  the  moment  triangle  or  polygon.  The  pin 
diameter  is  then  readily  found  in  the  following  manner. 

If  M  IS,  the  external  bending  moment,  /the  moment  of  in- 
ertia of  the  normal  section  of  the  pin  about  its  diameter  D 
and  K  the  intensity  of  stress  in  the  fibres  most  remote  from 
D,  then    it    is    known    from    the    theory   of   flexure,  since   1 

,  that 


64 


M: 


32 


D 


2.2  j/: 


M 
K 


(I)- 


SIZE   OF  PIXS.  349 

If  K  is  known,  Eq.  (i)  gives  D  at  once  after  M  is  found  by 
the  general  method  cxempHfied  by  Fig.  i,  or  in  any  other 
manner. 

For  wrought-iron  pins  in  the  trusses  of  railway  bridges,  K 
is  usually  taken  at  15,000  pounds.  This  value  in  Eq.  (i)  gives 
for  wrought-iron  pins  : 

D  =  0.089  VJT (2). 

For  steel  pins,  under  similar  conditions.  A' may  be  taken  at 
20,000  pounds,  for  which  : 

i;  =  0.081  \jJi (3). 

There  are  numerous  tables  showing  the  bending  moments 
of  pins  of  all  usual  diameters  with  given  values  of  A',  so  that 
in  practice  the  computations  expressed  in  Eqs.  (i),  (2)  and 
(3)  are  seldom  necessary.  The  value  of  M  is  determined  for 
any  particular  case,  after  which,  by  the  simple  inspection  of 
a  table,  the  proper  diameter  may  be  chosen. 

It  is  seen  by  Eq.  (i)  that  the  diameter  of  a  pin  varies  di- 
rectly as  the  cube  root  of  M  and  inversely  as  the  cube  root 
of  A'. 

It  may  sometimes  happen  that  Mt,  in  Fig.  i,  is  so  small 
that  it  may  be  neglected  ;  in  which  case  M  —  M' . 

No  pin  should  possess  a  diameter  less  than  eight-tenths  the 
width  of  the  widest  bar  coupled  to  it. 

When  bending  and  bearing  are  properly  provided  for,  a  safe 
shearing  resistance  will  be  amply  secured.  If  the  apparent 
moment  in  the  pin  is  sufficient  to  cause  failure  by  flexure,  it 
does  not,  by  any  means,  follow  that  failure  will  actually  take 
place;  for  the  distortion  of  the  pin  beyond  the  elastic  limit 
will  relieve  the  outside  eye-bars  of  a  larger  portion  (in  some 
cases  perhaps  all)  of  the  stress  in  them.  This  result  will  pro- 
duce a  redistribution  of  stress  in  the  eye-bars,  by  which  some 
will  be  understrained  and  the  others  correspondingly  over- 
strained. Thus,  although  the  pin  may  not  wholly  fail,  the 
safety  of  the  joint  will  be  sacrificed  by  the  overstrained  metal 
in  the  eye-bars. 


350  DETAILS  OF  CONSTRUCTION. 

Art.  75.— Camber. 

Camber  is.  the  curve  given  to  the  chords  of  a  bridge,  caus- 
ing the  centre  to  be  higher  than  the  ends,  or  rather  it  is  the 
amount  of  rise  of  the  centre  above  the  ends.  It  is  given  to  a 
truss  so  that  the  chords  may  not  fall  below  a  horizontal  line 
when  the  load  is  applied.  Fig.  8,  PI.  XII.,  represents  a  truss 
with  exaggerated  camber.  The  actual  amount  varies  from 
■^l^th  to  yaVoth  of  the  span. 

Camber  may  be  given  to  a  truss  either  by  lengthening  the 
upper  chord  or  shortening  the  lower  one ;  the  latter  method 
is  preferable  because  the  upper  chord  is  sometimes  not  hori- 
zontal, and  different  panel  lengths  would  have  to  be  shortened 
by  different  amounts. 

On  account  of  the  unavoidable  play  at  the  joints  of  all 
work,  the  shortening  of  the  lower  chord,  or  lengthening  of 
the  upper,  must  be  increased  by  about  ^V^  of  an  inch  per 
panel  in  order  to  secure  the  desired  camber. 

The  lower  chord  shortening  is  made  uniformly  throughout 
its  length  ;  that  is,  each  panel  length  is  shortened  by  a  con- 
stant quantity.  The  true  chords  will,  therefore,  become  arcs 
of  circles  of  very  large  radii,  and  vertical  posts  will  become 
radial. 

By  means  of  the  equation  of  the  circle,  y^  =  2Rx  —  x^,  R 
being  the  radius,  the  amount  of  shortening  or  lengthening  of 
chord  to  produce  a  given  camber  may  be  determined  if  the 
play  at  the  joints  be  omitted.  In  the  equation  above,  y  rep- 
resents the  half  span  and  x  the  camber  desired,  hence  the 
radius 


R  = 


__y'  +  x^ 


2X 


This  is   the   radius    of   the   lower  chord  when   cambered. 
Generally  it  will  be  near  enough  to  put  R  —  — 

2X 

The  angular  length  of  the  lower  chord  will  be 

1  J'  -1    2xy 

a—  2  sin-^  -f^  —  2  stn~^  -s — ^ , 
R  jr  +  x^ 


CAMBER.  351 

and  the  length  in  feet  : 

/  =  Ra. 

The  length  of  the  upper  chord  will  then  be : 

/■  =  {R^-  d)  a. 
The  difference  in  length  of  chords  will  be : 
D  ^  i  -  I  =  da. 

This  is  the  amount  by  which  the  lower  chord  is  to  be 
shortened,  or  the  upper  lengthened,  in  order  to  produce  the 
required  camber,  if  no  play  or  strains  exist. 

Since  x  is  very  small  compared  with  y  : 

a —  2  sur^  -^ — =^  =  -^p  ,      .,  =  ^^—  (nearly). 

If  the  span  is  l^ : 

S  X  ,     „       ^^^ 

a  =  —z—  ;  and,  //  =  -j— . 

h  n 

If  r  is  such  a  ratio  that  d  =  r/j : 

D  =  8rx. 

On  account  of  the  play  at  the  joints,  x  should  be  taken  a 
little  larger  than  the  camber  desired. 

Frequently  r  is  about  one-eighth,  and  for  such  a  value: 

D  =  x; 

or,  neglecting  the  play  at  the  joints,  t/ic  difference  in  lengths 
of  the  chords  should  equal  the  camber. 

If  the  chords  are  to  be  horizontal  under  the  greatest  loads, 
while  T  and  C  represent  the  supposed  uniform  intensities 
of  tension  and  compression  in  the  lower  and  upper  chords 
respectively,  E  and  E'  representing  the  coefficients  of  elas- 
ticity ; 

T         C 


352  DETAILS    OF  CONSTRUCTION. 

This  formula  can  only  be  approximate,  for  the  chords  are 
never  exactly  uniformly  stressed,  and  the  coefficient  of  elas- 
ticity is  probably  never  the  same  throughout  either  chord. 

Since  d,  the  depth  of  truss,  does  not  vary,  these  formulae 
apply  only  to  trusses  of  uniform  depth. 

A  "through"  truss  has  been  supposed,  but  the  same  for- 
mulae  exactly  apply  to  a  deck  bridge. 

It  is  to  be  borne  in  mind  that  one-half  the  horizontal  dis- 
tance between  the  centres  of  end  pins  is  to  be  taken  iox  y  in 
determining  R.  If  this  distance  is  assumed  in  designing  the 
truss,  then  the  panel  length  is  to  be  found  by  dividing  /  or  /' 
by  the  number  of  panels. 

If  the  panel  length  is  first  assumed,  and  the  camber  pro- 
duced by  shortening  or  lengthening  it,  then  this  horizontal 
distance  is  essentially  equal  to  the  assumed  chord  length 
diminished  or  increased  by  D  —  da. 

In  order  to  hold  the  camber  in  a  truss,  the  diagonals  must 
be  shortened,  as  shown  in  Fig.  9,  PI.  XII.  The  diagonal 
which  was  bd  before  cambering,  becomes  ed  afterward,  ad 
and  be  are  supposed  to  be  panels  in  the  upper  and  lower 
chords  respectively  before  putting  in  the  camber ;  afterward 
be  becomes  ef,  while  ad  remains  the  same  ;  the  lower  chord  is 
supposed  to  be  shortened.  Let  x  be  the  amount  of  shorten- 
ing  of  each  panel  of  the  lower  chord  ==  2be  =  2/e ;  d,  the 
depth  of  the  truss  ;  and/  the  original  panel  length  equal  to 
ad.     Then 


ed  =  ^d<?  +  ee"  =  i/  d-  +  [p -^ 


^d(?  4-  e^  =  A^ 


If  the  camber  is  produced  by  lengthening  the  upper  chord, 
then  ef'is  the  original  panel  length,  and  ad  the  new  one,  and 


ed=  ^d(?  -I-  e^=  i/ ^'  +  (/'  +  ")• 
In  a  triangular  truss  the  diagonal  ^e  Fig.  10,  is  changed  to 


ECONO^f/C  DEPTH  OF    TRUSSES.  353 

If  the  upper  chord  is  lengthened,  eg  is  the  diagonal  desired, 
send  ff  the  original  panel  length/.     Hence, 


cg=  V  d''  +  I  (/  +  x)\ 

Each  diagonal  is  to  be  shortened  to  the  length  cd. 

In  a  draw-bridge  each  arm,  in  giving  the  camber,  can  be 
considered  one  span,  but  the  whole  amount  of  shortening  in 
the  lower  chord  of  otte  arm  must  also  be  taken  out  of  the  up- 
per chord  at  the  centre.  If  this  is  not  done,  the  ends  will  sink 
below  their  original  positions. 

Art.  76. — Economic  Depth  of  Trusses  with  Parallel  Chords. 

The  so-called  economic  depth  of  truss  for  a  given  span,  is 
that  depth  which  involves  the  least  material  or  weight  of 
mjetal  in  the  bridge.  This  depth  depends  upon  the  intensity 
of  moving  load  for  each  truss,  the  length  of  panel,  the  great- 
est allowable  stresses,  etc.,  etc.  Various  mathematical  inves- 
tigations have  been  made  with  a  view  to  the  determination 
of  this  depth  of  truss  in  terms  of  the  length  of  span.  But 
on  account  of  the  exceedingly  intricate  character  of  the 
problem,  any  feasible  analysis  must  be  based  upon  assump- 
tions which  simplify  the  analytical  operations,  but  render  the 
results  only  approximately  true.  These  investigations,  how- 
ever, and  the  experience  of  American  engineers,  show  that  a 
depth  varying  from  one-fifth  to  one-seventh  the  length  of 
span  will  give  the  least  weight  of  truss ;  the  former  for  very 
heavy  loads,  as  in  two  truss  double  track  bridges,  and  the 
latter  for  light  loads. 

When  the  span  becomes  very  long,  /".  r.,  400  to  500  feet, 
the  depth  of  truss  increases  to  an  unusual  height,  and  the 
cost  of  erection  is  correspondingly  large.  The  depth  is  then 
frequently  taken  not  larger  than  one-eighth  the  span,  or  even 
less. 

Again,  local  conditions,  such  as  the  necessarily  uniform 
depth  (for  the  sake  of  appearance)  of  adjacent  spans  of  vary- 
ing length,  sufficient  depth  of  short  spans  for  over-head  brac- 
ing (very  necessary  for  lateral  stability),  etc.,  in  the  majority 

3i 


354  details;  of  construction. 

of  cases  exclude  the  use  of  the  economic  depth,  even  if  it 
were  exactly  known. 

It  is  to  be  borne  in  mind,  also,  that  the  lightest  truss  is  not 
necessarily  the  cheapest.  That  bridge  is  the  most  economi- 
cal which  can  be  made  ready  for  traf^c  for  the  least  money. 

Facility  in  working  up  details,  and  the  least  possible  amount 
of  time  in  the  shop,  are  very  important  elements,  indeed,  in 
every  design. 

In  fact,  the  lightest  weight  does  not  make  the  most  econom- 
ical bridge,  for  the  reason  that  the  shop  cost  per  pound  is 
greater  than  with  a  somewhat  increased  weight  of  metal. 
When  it  is  borne  in  mind  that  a  considerable  variation  may 
be  made  from  the  depth  of  least  weight,  without  affecting 
that  weight  to  any  considerable  extent  (as  actual  computa- 
tions show  to  be  the  case),  it  is  easy  to  understand  that  the 
truly  economic  depth  is  materially  less  than  that  which  gives 
precisely  the  least  weight  of  material. 

Long  panels  are  an  economic  feature  of  any  bridge  possess- 
ing a  system  of  floor-beams  and  stringers,  as  well  as  condu- 
cive to  other  points  of  merit.  The  resulting  concentration  o*^ 
m.etal  not  only  leads  to  less  weight  and  rate  of  cost  in  the 
shop,  but  enhances,  also,  the  stiffness  and  stability  of  the  in- 
dividual members. 

For  economy  in  weight,  long  panels  require  a  greater  depth 
than  shorter  panels. 

This  much  may  be  said  in  regard  to  continuous  trusses: 
On  account  of  the  existence  of  the  points  of  contrailexure, 
they  require  considerably  less  depth  than  trusses  that  are 
not  continuous,  used  on  the  same  points  of  support.  The 
depth  of  the  latter,  therefore,  is  a  limit  which  should  never 
be  reached  by  the  depth  of  the  former. 

Art.  77.— Fixed  and  Moving  Loads. 

Both  fixed  and  moving  loads  depend  upon  the  local  cir- 
cumstances of  each  case,  and  the  former  very  much  upon  the 
character  of  the  design.  A  depth  from  one-fifth  to  one- 
seventh  the  span  will  give   a  very  light  fixed    weight,  but  a 


FIXED  AND   MOVING  LOADS.  355 

depth  of  one-twelfth  the  span  will  involve  a  considerable  in- 
crease of  weight,  while  the  moving  load  remains  the  same. 

The  weight  of  a  single-track  railway  floor,  for  the  present 
(1885)  existing  moving  loads,  may  be  taken  at  about  400 
pounds  per  foot. 

The  moving  load,  also,  depends  upon  the  length  of  span. 
If  the  span  is  very  great,  the  probability  of  the  whole  bridge 
being  covered  with  an  excessively  heavy  moving  load  is  very 
slight,  if  any  exists  at  all.  If  the  span  is  short,  one  or  two 
locomotives  may  cover  the  whole  bridge,  thus  causing  the 
moving  load,  per  foot,  to  be  very  great  for  the  whole  span: 

Thus  it  is  seen  that  the  moving  load,  per  foot,  may  decrease 
as  the  span  increases. 

The  whole  matter  of  moving  loads  for  both  highway  and 
railway  bridges  is  well  illustrated  by  the  following  tables, 
taken  from  '*  A  Bill  to  secure  greater  Safety  for  Public  Travel 
over  Bridges,"  introduced  in  the  Sixty-second  General  As- 
sembly of  the  State  of  Ohio,  shortly  after  the  Ashtabula 
■disaster : 

For  City  and  Suburban  Highway  Bridges. 

span  in  feet-  Moving  load  per  square  foot. 

O     to       30 1 10  pounds. 

30         "  50 100 

50  "       75 90       " 

75  "     100 80 

100  "     200 75 

2(X)  "     400 65       " 

All  other  Highway  Bridges. 

span  in  feet.  Moving  load  per  square  foot. 

o  to   30 100  pounds. 

30  "   50 •  90 

50  "   75 80   " 

75  "  100 75 

100  "  200 60 

200  "  400  50 


356 


DETAILS   OF  CONSTRUCTION. 


Railway  Bridges. 


Moving  load  per  lineal 
foot  of  each  track. 


Span  injeet. 

o  to      7^ 9,000  pounds. 

1\  "    10  7,500 

10  "       \2\ 6,700 

\2\  "       15     6,000 

15  "      20    5,000 

20  "       30    4,300 

.30  "      40    3700 

40  "     50 3'300 

50  "     75    3>200 

75  "  100   3,100 

100  "   1 50   3,000 

1 50  "  200    2,goo 

200  "  300   2,800 

300  "  400   2,700 

400  "   500   2,500 


Floor-beams  and  stringers  are  really  bridges  of  short 
spans  equal  to  their  lengths,  consequently  they  must  be 
designed  for  the  heavy  loads  belonging  to  those  short 
spans.      Fig.    i   shows   the    locomotive    weight    specified    by 


Fig.  I. 


Mr.  Theodore  Cooper,  C.E.,  in  his  "  General  Specifica- 
tions for  Iron  Bridges  and  Viaducts,"  while  Fig.  2  shows 
the  heavy  passenger  locomotive  used  by  Mr.  Jas.  M.  Wil- 
son, C.E.,  in  his  standard  specifications  for  the  Pennsyl- 
vania R.  R. 


FIXED  AND   MOVING   LOADS. 


357 


These  represent  the  heaviest  engines  of  their  type  now  in 
use.  There  is,  however,  a  heavy  decapod  engine  shown  by 
Fig.  3,  beginning  to  make  its  appearance. 


K:-5-'0'-^^-5'6--^ — 9'0--^— 8'0-— ►]. 9'6-— *)— 5'0-4^5'6-4"5V4-3^ij 


^-»J 


Fig.  2. 

The  moving  load  usually  specified  consists  of  two  of  any  of 
these  types  of  locomotives  followed  by  a  uniform  train  of 
3,000  pounds  per  lineal  foot.  Occasionally  the  locomotive 
concentrations  are  followed  by  those  of  the  train. 

Besides  the  preceding  heavy  moving  loads,  there  are  in- 


Km. 


A    si 


s'    81     «l    SI 

I  v..  .',•  1' .0.  Li 


I  '  ,r  '  I  '  '  '      I 


-M'S- 


FiG.  3. 


numerable  lighter  ones  depending  upon  local  circumstances 
of  trafific. 

The  actual  concentrations  play  a  very  important  part  in 
bridge  computations.  The  old  method  of  a  uniform  load, 
even  with  an  engine  excess,  no  longer  fulfils  the  requirements 
of  the  best  engineering  practice,  particularly  in  the  treatment 
of  short  spans. 

This  is  well  illustrated  by  the  following  table  which  shows 
the  uniform  load  per  lineal  foot  which  will  produce  the  same 


358  DETAILS   OF  CONSTRUCTION. 

chord  stresses  at  the  centre  of  the  span  as  the  actual  concen- 
trations shown  in  Fig.  i. 


span  in  feet. 

Equ 

iv.  uniform 

:  load 

Span 

'.  in  feet. 

Equiv.  uniform  load 

in 

lbs.  per  lin. 

ft- 

in 

lbs.  per  lin.  .ft. 

55 

3750 

25 

4,838 

50 

3,866 

20 

5,137 

45 

4,004 

15 

5,760 

40 

4,242 

12 

6,000 

35 

4-336 

10 

5,766 

30 

4,572 

5 

9,600 

Above  55  feet  the  equivalent  uniform  load  per  lineal  foot 
will  slowly  decrease  until  it  reaches  a  value  of  about  3,200 
lbs.  for  100  feet  and  over,  /.  r.,  supposing  the  moving  load  to 
consist  of  a  train  of  such  locomotives. 


Art.  78. — Safety  Factors  and  Working  Stresses. 

Although  the  subjects  of  safety  factors  and  working 
stresses  properly  belong  to  the  domain  of  the  resistance  of 
materials,  they  may  here  be  touched  upon  in  a  general 
manner. 

The  fixed  weight  of  a  long  span  bridge  is  much  greater, 
per  foot,  than  that  of  a  short  span.  Again,  it  has  been  seen 
in  the  preceding  article  that  the  moving  load  for  a  long  span 
is  much  less  than  that  for  a  short  span.  For  both  these  rea- 
sons, the  variations  of  stress  in  passing  from  a  loaded  to  an 
unloaded  condition  are  much  greater  in  the  material  of  a 
short  span  than  that  of  a  long  one.  Consequently,  the  mate- 
rial will  be  much  more  fatigued  in  a  short  span  than  in  a 
long  one. 

Although  the  subject  of  the  fatigue  of  metals  is  yet  in  an 
unsettled  state,  it  is  clearly  established  that  these  conditions 
of  stress  in  short  spans  demand  a  larger  safety  factor,  or 
smaller  working  stress,  than  those  in  the  long  spans. 

Again,  in  any  bridge  or  truss  whatever,  carrying  a  moving 
load,  some  parts  are  subject  to  a  much  greater  variation   of 


SAFETY   FACTORS  AND    WORKING    STRESSES  359 

stress  in  the  process  of  first  bein<j  subject  to,  and  then  relieved 
of,  loads  than  others. 

Counter-braces  may  not  be,  and  probably  are  not,  strained 
at  all  by  the  fixed  load  ;  but  they  take  a  proper  working  stress 
under  the  action  of  the  moving  load. 

The  condition  of  loading  for  greatest  stress  in  any  main 
web  member,  except  those  at  the  ends,  is  a  partial  covering 
of  the  span.  But  the  fixed  load  is  distributed  over  the  whole 
span.  Hence  the  variation  of  stress  in  the  main  web  mem- 
bers will  be  greatest  at  the  middle  of  the  span,  and  least  at 
the  end.  At  the  centre,  however,  the  variation  is  much  less 
than  in  the  counter-braces. 

The  fatigue  of  the  material,  therefore,  requires  that  the 
greatest  safety  factors,  or  least  zuorking  stresses^  be  found  in  the 
counter-braces ;  and  that  the  ivorking  stresses  in  the  main  zveb 
members  at  the  centre  be  greater  tJian  those  in  the  counter-braces, 
but  less  tJian  those  in  the  main  zveb  members  at  the  ends  of  the 
truss. 

The  disposition  of  the  moving  load  for  the  greatest  chord 
stresses  is,  in  all  cases,  essentially  the  same  as  that  of  the  fixed 
load.  Hence  the  variation  of  stress  will  be  essentially  the 
same  throughout  the  chords,  and  the  safety  factor  or  working 
stress  may  be  uniform  throughout  each  chord  ;  the  safety 
factor  being  the  same  as  that  in  the  end  web  members  sus- 
taining the  same  kind  of  stress. 

If  a  structure  is  to  carry  a  fixed  load  only,  the  safety  factor 
may  be  three  for  wrought-iron  and  steel,  and  possibly  as  small 
for  good  qualities  of  cast-iron  and  timber.  As  a  rule,  how- 
ever, cast-iron  and  timber  require  a  larger  safety  factor  than 
wrought-iron  and  steel.  Local  circumstances  affect,  to  a 
great  extent,  working  stress.  If  the  risk  (respecting  life  and 
property)  attending  failure  is  small,  the  safety  factor  may  be 
small  also.  But  if  the  risk  is  great,  the  safety  factor  must  be 
correspondingly  great. 

In  the  truss  members  of  long  span  bridges  of  wrought-iron 
and  steel,  the  safety  factors  may  vary  from  three  and  a  half 
or  four  to  five  ;  but  in  short  spans  of  the  same  material,  they 
should  vary  from  about  five  to  six  or  eight. 


360  DETAILS  OF  CONSTRUCTION. 

Good  cast-iron  should  be  found  with  safety  factors  varying 
from  six  to  ten,  while  those  for  timber  may  vary  from  eight 
to  twelve. 

It  is  not  to  be  supposed  from  these  large  safety  factors  that 
the  determination  of  the  stresses  or  the  character  of  the  vari- 
ous materials  is  so  excessively  uncertain.  It  is  certainly  true 
that  there  is  some  indetermination  in  these  respects,  but  only 
a  little  in  comparison  with  that  connected  with  tJie  mode  of 
applicatioji  of  the  moving  load. 

'  With  a  perfect  condition  of  track,  a  rapidly  moving  train  is 
supposed  by  many  to  approximate  ver>'  closely  to  a  suddenly 
applied  load,  although  it  is  quite  certain  that  it  does  not. 
For  this  reason  some  engineers  have  doubled  the  moving 
loads,  in  making  their  calculations,  and  then  fixed  the  values 
of  the  safety  factors  as  if  all  loads  were  gradually  applied. 

But  no  track  is  in  perfect  condition,  and  all  rough  places, 
or  lack  of  continuity,  such  as  rail  joints  more  or  less  open, 
produce  shocks  which  cause  greater  stress  than  any  suddenly 
applied  loads.  The  amounts  of  these  last  stresses  are  inde- 
terminate, for  the  extent  of  their  causes  can  scarcely  be 
determined. 

Again,  Mr.  J.  W.  Cloud,  C.  E.,  at  the  Philadelphia  meeting 
of  the  American  Institute  of  Mining  Engineers,  February, 
1881,  pointed  out  the  existence  of  certain  unrecognized 
stresses ;  such  as  those  caused  by  the  vertical  component  of 
the  thrust  of  the  connecting-rod  of  a  locomotive,  which  alter- 
nates in  direction  twice  in  each  revolution  of  the  driving- 
wheels,  thus  producing  a  pulsating  effect,  as  well  as  those 
which  arise  from  the  lack  of  balance  of  the  driving-wheels  in 
a  vertical  direction. 

All  these  causes  produce  stresses  which  it  is  impossible  to 
measure,  and  the  safety  factor  must  cover  all  uncertainties. 

It  is  possible  that  a  more  highly  perfected  track  and  the 
production  of  more  nearly  uniform  material  in  connection 
with  an  extended  experience  may  justify  the  reduction  of 
safety  factors. 

The  following  "  Table  of  Tubular  and  Truss  Bridges  for 
Single  and  Double  Track  Railways,  constructed  of   Iron  and 


SAFETY  FACTORS  AND    WORKING    STRESSES.         36 1 

Steel  and  having  Spans  exceeding  300  feet,"  gives  the  work- 
ing stresses,  loads,  and  other  interesting  data  of  some  of  the 
principal  bridges  of  the  world.  It  is  taken  (with  the  excep- 
tion of  No.  18)  from  the  "  Proceedings  of  the  Institution  of 
Civil  Engineers"  of  Great  Britain,  Vol.  LIV.  The  greater 
portion  of  it  is  there  given  in  connection  with  a  paper  by 
Mr.  T.  C.  Clarke,  on  "  Long  Span  Bridges." 


No. 

Date  of  Erection. 


?; 

O 

u 

^'i- 

n  C 

C/i_ 

-    o 

^ 

>^ 

(/)_, 

>? 

-  r 

6= 

•    u> 

■    D 

:  a. 

2.  IT 


Sr 


o 


ffic 
m  ft 

03 

ft 
-     n 


Span  in  Feet  between  Points  of 
Bearing^ 


Tons  of  Iron.    (2000.00  lbs.) 


•^     Width  between  Centres 
Trusses. 


of 


.''B 


4, 

„ 

u. 

-P' 

» 

4. 

-u 

n 

Total  Dead  Load  of  Iron 

Tia  0 

a\ 

8 

o\ 

^ 

0 

0 

8 

§ 

0 

and  Timber  in  Track. 

0      0 

» 

K> 

» 

0 

w 

Live  Load  of  Engines 

8 

^ 

8 

■^ 

8 

0 

8 

4>. 
0 

and  Cars. 

a) 

5     1 

^j 

8, 

-^ 

■^ 

S 

^0 

'^ 

1 

From  Fixed  Load  only. 

H 

0 

0 

0 

Ul 

M 

' 

n 

3- 
0 

8 

oFo 
o'S  0 

n 

0 

1 

R 

8 
R 

0 

From  Total  Fixed  and 
Moving  Loads. 

s 

5 

0  tfl  0 

Q. 

0 

to 

0 

B 

4- 
0 

8 

en 

0 

8 

"8 

From  Fixed  Load 
only 

if 

1 

I 

^ 

0° 

00 

c^ 

>o 

From  Total  Fixed 

§ 

§ 

8 

§ 

8 

and  Moving   Loads. 

m        z 

» 

*»j 

£ 

0 

■u 

Test  Load.    Tons.    (2000  lbs.) 

- 

- 

- 

0 

" 

^^ 

^ 

? 

0 
0 

^j 

-vl 

P 

Centre  Deflection. 

00 

CO 

H 

0 

OJ 

i 

Dead  Load  of  Iron,  Tin 
Deflection  of  Span  from 

iber,  &c. 

1^ 

^ 

its  own 

Ln 

p 

Weight  on  Removal  of  Scaffold'g. 

H 

> 

H 

;? 

> 

H 

H 

> 

0  ^ 

s*"  S 

c 

r*  — 

0 

^ 

angular  truss 
3  cliords  cast 
ailed      iron, 
onnections ; 
angular  truss 

0 

il 

0 

3  :q 
J?  0 

3 

P  0  0T3 

"■  s  - 

0. 
0 

3 

1x5 

Si 
'  0 

c 

S3 

wo  P  0 

2-33 
7  "TO  3 

0  g  ETS 
Sag- 

5 
cr 

3 

a 

3 
3 

g) 

J3 

St 

r 

3  S 

c 

Tl 

? 

1:1 

•    X) 

p  t 

^5 

'!£ 

V-3 

^5 

3 

c 

Z 

3 

1 

—  o"'3uOM--:i 


ii  o    . 
—  ■-  2 


-^  _—  o  "1  ^ 

'  3  3  ■»  "  '-•T  ' 


Qli  rt  w 

".^ 

^  n; 

C  ,„  3  S     , 

T3  rt 

ed   ir 
ction 
lartr 
ca 
ctual 

=  w  3  "O  nj 

a 

w  C  C  rt     . 

=  ii 

0 

—  u  »i  b.s 

< 

<5 

8 

R 

8 

o 

8 

o 

M 

OO 

o 

0 

o 

o 

g 

OO 

VO 

S 

o 

S 

0 

8 

0 

8 

8 

vC 

NO 

ir> 

" 

I^ 

£? 

m 

Di  ^ 


-i     Oi 


■£0 
O 


'rt 

V 

rt 

irt 

^ 

3 

•3 
c 

rt 

u 

.yi 

S 

J3ffi 


'  OvO 


OO 

0 

0 

0 

0 

0 

s 

•^ 

On 

:r 

::• 

lO 

c 

o 

in 

o 

in 

0 

o 

OO 

o 

r» 

o 

t^ 

8 

« 

0 

s 

0 

Ov 

M 

"' 

'^ 

xn 

W^'  ■ 

.    £E  «^ 

•'^     i  B  '.' 

3  -l.s 


Q 

_< 

Q 

— 

M 

Q 

— 

^^ 

— 

S 

'-' 

00                                             no' 
r'N                                                 ON 

o 

r-^ 

2 

t^oo 

NO      1-^ 

2S 

00 

00     vo    r^ 


13 

-^ 

o 

o  m 

f*^ 

m 

t^ 

8 

C4 

^  <4- 

■«• 

■a- 

^  N 

— 







~ 

in 

0 

Ul 

m 

• 

00* 

OO 

I^NO 

f-. 

c* 

CI 

« 

NO 

00 

0 

c^ 

■+ 

N 

m 

tN. 

r^ 

ON 

U^OO 

-* 

0^ 

O 

r^ 

^ 

r- 

Jl 

m 

rr> 

2l 


(55      oS 


J  ^"C  13 


O      3  S     U 


"a  'CC    g 

>  WW  2 


s. 

■^ 

3n 

- 

I!          J^    M 

:; 

0 

NO 

No. 

1 

s 

» 

00 

OO         OD          00  OO         00 

s. 

Date  of  Erection. 

0 

^ 

00         0           0    0          4. 

0 

KJ> 

2 

1 

< 

n 

X 

»      JO      ?3H      O 

■<" 

w 

-  3 
2.0 
<!  £ 

22. 

2  3 

'S-as-— s'S's-H- 
:  1  <-<i--^' 

u 
re 

o 

ST 

re 

ft 

.    a- 

D3 
o 
3 
3 
2. 

■    a 

n 

•  -o 

.  n 

o 

"  ■   2> 

2_ 

re 

n 

< 

o 

3- 
3 

W 

:  S 

< 

■-t 

■  u-      -  ■    C?'< 

■  "  ■    S  •   2  ° 

N    ■      T  •      ft   ■ 

re' 

?5 

c 

o 

O 

O 

> 

rn   5   3)   T^K 

Q 

t/i 

< 

n 

^   1   1   P 

W 

2 
o 
2, 

3 
d 

3 
3 

re 
3) 
re 
3 

3 

•s. 

5' 

re 

0 
3 

n 

3 

0 
3 

n 

j:. 

OJ 

l<j       u;        w       OJ  U) 

,^ 

,„ 

Span  in  Feet  between  Points  of 

0 

Co 

M 

Bearing. 

■t- 

VO 

2 

VI         OJ         Oi  Ln 

OJ         so          0  i-n 
00       4.          On  to 

^ 

Cn 

Tons  of  Iron.     (2000.00  lbs.) 

? 

r^ 

Ct\ 

00       oi          ON'-n 

3s 

r^ 

yi 

Width  between  Centres  of 

"- 

N) 

"-" 

OJ 

OJ 

3 

Trusses. 

a> 

Vl 

^ 

^         OJ          0    0 

o-^ 

0 

J" 

z 

d 
s 

z 

0 

u 

„ 

..          «          „ 

^ 

re 

3 

8 

~ 

On 

4.          OJ           0  4. 

^ 

CO 

P 

IW 

z 

1) 

n 

, 

? 

u\ 

-^i     W 

-J    - 

4.    M  4.             OJ    M 

4^    ro 

M 

4.     N  NO     0       4^   ON 

O  OJ 

^^ 

w 

0 

-J  0 

OD  00 

Ln-J  OJ    0        M  OJ 

3 

3- 

' 

' 

Ln 

^ 

<^ 

4.        ro        K)  OJ 
N         4.         NO  +■ 

^ 

* 

Total  Dead  Load  of  Iron 

?Fs 

0 

N 

Jo 

NO         <Ji          OOOJ 

ul 

-O 

and  Timber  m  Track. 

O    0 

VO 

!o 

t      ii      «  - 

w 

"oJ 

Live  Load  of  Engines 

-i  p  "^      1 

o  o 

U( 

" 

M           ON         M    0 

^ 

t 

and  Cars. 

X 

z    1 

Ov 

en 

on 

ON 

«) 

^ 

4k 

§ 

From  Fixed  Load  only. 

'     ■"■ 

r..             ..            M     -• 

Z 

n 

•ji    0 

c2       o"       iZ° 

NO 

From  Total  Fixed  and 

'£ 

- 

88 

OJ          CO        OOOJ 

NO 

Movmg  Loads. 

a 

z 

TJ 

n 

On 

0 

On 

From  b'lxed  Load 

•n 

NO 

O 

only. 

So 

50 

0 

3^:;; 

^  2 

CO 

3            -             "     0 

NO 

From  Total  Fixed  and 

?*  8 

4>- 

Moving  Loads.          ? 

z 

OJ 

4».    M 
It 

'oJ 

Test  Load.    Tons.    (2000  lbs.) 

„ 

„ 

«    « 

^ 

i__^ 

OJ 

Ui 

OJ  '!n 

•^ 

3 

Centre  Deflection. 

fj 

OO'jJ 

ON 

Ol 

VO 

UN 

, 

H 

4^ 

■^ 

^ 

o 

Dead  Load  of  Iron,  Timber,  &c. 

Deflection  of  Span  from  its  own 

3 

Weight  on  Removal  of  Scaffold'g. 

u> 

O                         t/! 

D 

■0  t/)3 

0                      ,3 

0 

o  sriw 

C                       'TQ 

sS;r 

s           s- 

cr 

«!    — 

"                « 

re 

W 

2  3 

-1                   ;p 

2 

S  w 

KT 

?r 

w 

T3  ■ 

■0 

3E 

GENERAL   OBSERVATIONS.  365 

Art.  79. — General  Observations. 

All  abutting  surfaces  in  bridges,  or  similar  structures, 
should  be  very  carefully  machine  finished. 

Where  pins  bear  against  portions  of  the  upper  chord,  as  at 
c,  d,  and  e  of  Fig.  5,  PI.  III.,  the  amount  of  bearing  surface 
should  be  determined  as  for  rivets,  and  sufficient  area  given 
by  riveting  on  thickening  plates,  if  necessary.  A  thickening 
plate  is  shown  at  the  joints  of  the  same  figure.  The  num- 
ber of  rivets  for  the  thickening  plate  is  determined  by  the 
amount  of  pressure  allowed  on  each  one,  as  has  already  been 
shown. 

If  a  finished  piece  is  to  fit  into  a  finished  cavity,  however 
well  the  work  may  be  done,  there  must  be  at  least  ^^  inch 
^'  play." 

One  end  of  a  truss  bridge,  unless  the  span  is  very  short, 
usually  rests  upon  "  expansion "  rollers,  from  two  to  four 
inches  in  diameter.  An  approximate  formula  for  the  resist 
ance  of  such  rollers  is  given  in  the  Appendix. 


CHAPTER  XI. 

WIND    STRESSES   AND    BRACED   PIERS. 
Art.  80.— Wind   Pressure. 

In  a  paper  presented  to  the  American  Society  of  Civk 
Engineers  (Transactions,  Vol.  X.),  Mr.  C.  Schaler  Smith  gives 
some  very  valuable  information  in  regard  to  wind  pressure. 
The  highest  observed  pressure  which  has  come  within  his 
knowledge  is  93  pounds  per  square  foot.  This  pressure 
derailed  a  locomotive  at  East  St.  Louis,  Mo.,  in  1871.  In  his- 
own  specification  he  says  : — 

"  The  portal,  vertical,  and  horizontal  bracing  shall  be  pro- 
portioned for  a  wind  pressure  of  30  pounds  per  square  foot 
on  the  surface  of  a  train  averaging  10  square  feet  per  lineal 
foot,  and  on  twice  the  vertical  surface  of  one  truss." 

The  wind  pressure  on  a  train  is  a  moving  load,  and  should 
be  so  considered,  while  the  wind  pressure  on  the  trusses  is  a 
fixed  load. 

His  experiments  on  the  Rock  Island  draw-bridge  showed 
that  the  wind  pressure  against  the  two  trusses  was  over  1.8 
times  that  on  the  exposed  surface  of  one. 

Again,  quoting  from  his  paper : — 

"  The  Erie  specifications  are  as  follow : 

Fixed  load,  roadway  chord,  150  lbs.  per  lineal  foot. 

other  "        150    " 

Moving "       roadway      "        300     "       "        "         " 
Iron  in  tension  at  15,000  pounds. 
"     "  compression,  factor  4. 

"The  Pittsburg-,  Cincinnati  and  St.  Louis  Railway  requires 

366 


WIND   PRESSURE.  367 

300  pounds  per  foot  for  the  train,  and  30  pounds  per  square 
foot  on  one  truss  only. 

"  For  the  bridge  over  the  Missouri,  at  Glasgow,  50  pounds 
per  square  foot  on  one  truss,  and  300  pounds  per  lineal  foot 
of  train  were  used. 

"  For  the  Eads  bridge,  at  St.  Louis,  50  pounds  per  square 
foot  on  the  structure  alone  was  the  specified  pressure. 

"  For  the  Kentucky  River  bridge  the  wind  pressure  was 
assumed  at  314  pounds  per  square  foot  on  spans,  train,  and 
piers,  and  factor  4  was  used  in  proportioning  the  bracing. 

"  The  Portage  bridge,  New  York,  was  built  to  resist  30  pounds 
per  square  foot  on  structure  and  train,  and  50  pounds  per 
square  foot  on  the  structure  alone. 

"  The  520  feet  span  over  the  Ohio,  at  Cincinnati,  was  de- 
signed to  withstand  50  pounds  per  square  foot  on  structure 
alone,  or  30  pounds  per  square  foot  on  train  and  structure 
combined. 

"  A  fully  loaded  passenger  train,  and  the  heaviest  possible 
freight  train,  will  leave  the  track  at  the  respective  pressures  of 
31 4  and   56^  pounds  per  square  foot." 

Engineers  frequently  specify  30  pounds  per  square  foot  of 
trusses  and  train  combined,  or  50  pounds  per  square  foot  of 
trusses  alone. 

300  pounds  per  linear  foot  of  single  track  is  also  frequently 
used  for  moving  wind  pressure  on  train. 

The  following  refers  to  the  single  track  bridge  at  Platts- 
mouth,  Neb.,  and  is  from  the  Railroad  Gazette,  17th  Dec, 
1880:  "The  structure  is  also  designed  to  resist  a  lateral  wind 
pressure  of  500  pounds  per  lineal  foot  on  the  floor,  and  200 
pounds  per  lineal  foot  on  the  top  chord  of  the  through  spans 
and  the  bottom  chord  of  the  deck  spans;  these  quantities  are 
about  equivalent  to  a  wind  pressure  of  30  pounds  per  square 
foot  on  the  bridge  when  covered  by  a  train,  and  to  50  pounds 
per  square  foot  on  the  empty  bridge." 

The  following  are  a  set  of  rules  recommended  for  English 
practice,  almost  exactly  in  the  words  of  the  report : 


368  WIND   STRESSES  AND   BRACED  PIERS. 

Report  of  the  Committee  appointed  to  eoisidcr  the  Question 
of  Wind  PressKre  on  Raihvay  Striiet2ircs,  by  the  Board  of 
Trade  of  London,  made  on  the  20th  May,  1881. 

The  following  rules  were  recommended  : 

( i).  For  railway  bridges  and  viaducts,  a  maximum  pressure 
of  56  pounds  per  square  foot  should  be  assumed  for  purposes 
of  calculation. 

(2).  That  when  the  bridge  or  viaduct  is  formed  of  close 
girders,  and  the  tops  of  such  girders  are  as  high  or  higher 
than  the  tops  of  passing  trains,  the  total  wind  pressure  upon 
such  bridge  or  vi-aduct  should  be  ascertained  by  applying  the 
full  pressure  of  56  pounds  per  square  foot  to  the  entire  verti- 
cal surface  of  one  main  girder  only.  But  if  the  top  of  a 
train  passing  over  the  bridge  is  higher  than  the  tops  of  the 
main  girders,  the  total  wind  pressure  upon  such  bridge  or 
viaduct  should  be  ascertained  by  applying  the  full  pressure 
of  56  pounds  per  square  foot  to  the  entire  vertical  surface, 
from  the  bottom  of  the  main  girders  to  the  top  of  the  train 
passing  over  the  bridge. 

(3).  That  when  the  bridge  is  of  the  lattice  form,  or  of 
open  construction,  the  wind  pressure  upon  the  outward  or 
windward  girder  should  be  ascertained  by  applying  the  full 
pressure  of  56  pounds  per  square  foot,  as  if  the  girder  were 
a  close  one,  from  the  level  of  rails  to  the  top  of  the  train 
passing  the  bridge  or  viaduct,  and  by  applying,  in  addition, 
the  full  pressure  of  56  pounds  per  square  foot  to  the  as- 
certained vertical  area  of  surface  of  the  iron  work  of  the 
same  girder,  situated  below  the  level  of  the  rails  or  above 
the  top  of  a  train  passing  over  such  bridge  or  viaduct.  The 
wind  pressure  upon  the  inward  or  leeward  girder  or  girders 
should  be  ascertained  by  applying  a  pressure  per  square  foot 
to  the  ascertained  vertical  area  of  the  surface  of  the  iron 
work  of  one  girder  only,  situated  below  the  level  of  the  rails, 
or  above  the  top  of  a  train  passing  over  the  said  bridge  or 
viaduct,  according  to  the  following  scale  : 

{a).  If  the  surface  area  of  the  open  spaces  does  not 
exceed  f  of  the  whole  area  included  within  the  outline  of  the 


WIND    PRESSURE. 


369 


girder,  the  pressure  should  be  taken  at  28  pounds  per  square 
foot. 

{b).  If  the  surface  area  of  the  open  spaces  lie  between  | 
and  I  of  the  whole  area  included  within  the  outline  of  the 
girder,  the  pressure  should  be  taken  at  42  pounds  per  square 
foot. 

{c\  If  the  surface  area  of  the  open  spaces  be  greater  than 
I  of  the  whole  area  included  within  the  outline  of  the  girder, 
the  pressure  should  be  taken  at  56  pounds  per  square  foot. 

(4).  That  the  pressure  upon  arches  and  piers  of  bridges 
and  viaducts  should  be  ascertained,  as  nearly  as  possible,  in 
conformity  with  the  rules  above  stated. 

(5).  That  in  order  to  insure  a  proper  margin  of  safety  for 
bridges  and  viaducts,  in  respect  of  the  strains  caused  by  wind 
pressure,  they  should  be  made  of  sufficient  strength  to  with- 
stand a  strain  of  4  times  the  amount  due  to  the  pressure 
calculated  by  the  foregoing  rules.  And  that  for  cases  where 
the  tendency  of  the  wind  to  overthrow  structures  is  counter- 
balanced   by   gravity   alone,    a   safety    factor    of   2    will    be 

sufficient. 

John  Hawkshaw. 

W.  G.  Armstrong. 

W.  H.  Barlow. 

G.  G.  Stokes. 

W.    YOLLAND. 

The  evidence  before  us  does  not  enable  us  to  judge  of  the 
lateral  extent  of  the  extreme  high  pressures  occasionally  re- 
corded by  anemometers,  and  we  think  it  desirable  th:t 
experiments  should  be  made  to  determine  this  question.  If 
the  lateral  extent  of  exceptionally  heavy  gusts  should  prove 
to  be  very  small,  it  would  become  a  question  whether  some 
relaxation   might  not  be  permitted  in  the   requirements  of 

this  report. 

W.  G.  Armstrong. 

G.  G.  Stokes." 

Up  to  the  date  of  the  above  report,  the  highest  pressure 
per  square  foot  ever  recorded  at   Glasgow  was  47  pounds  ; 
24 


370 


WIND    STRESSES  AND   BRACED  PIERS. 


while  the  highest  ever  recorded  at  Bidston,  near  Liverpool, 
was  90  pounds  per  square  foot.  At  another  time,  at  Bidston, 
80  pounds  per  square  foot  was  recorded. 

The  above  committee  also  found  that,  if  P  is  the  maxi- 
mum pressure  per  square  foot,  Fthe  maximum  run  of  wind 
in  miles  per  hour,  both  these  quantities  being  observed 
by  anemometers,  the  following  equation  very  nearly  held 
true  : 


100 


Art.  81. — Sway  Bracing. 

The  construction  of  the  upper  and  lower  sway  bracing  of 
a  truss  must,  so  far  as  the  jar  and  oscillation  of  a  moving 
road  are  concerned,  be  a  matter  of  judgment ;  but  the  stresses 
due  to  the  action  of  the  wind  may  be  determined  with  suffi- 
cient accuracy. 

Although  a  moving  train  will  partially  shelter  one  truss,  it 
seems  no  more  than  prudent,  with  the  ordinary  open  style  of 
American  bridge,  to  consider  the  action  of  the  wind  as  exist- 
ing constantly,  during  the  passing  of  a  train,  over  the  whole 
of  the  projection  of  each  truss  in  the  bridge  on  a  plane  nor- 
mal to  the  direction  of  the  wind.  If  this  is  considered  exces- 
sive, however,  for  low  trusses,  that  portion  of  the  windward 
truss  sheltered  by  the  train  may  be  omitted. 

Let  Fig.  I  and  Fig.  2  represent  a  single  concellation  rail- 
way truss  bridge,  with  vertical  and  diagonal  bracing,  and 
let  the  wind  be  supposed  to  blow  in  the  direction  shown 
by  the  arrow,  which  is  normal  to  the  planes  of  the  trusses. 
Primed  letters  belong  to  the  truss  DC'N'O,'  but  all  are  not 
shown. 

As  the  truss  is  a  "  through  "  one,  all  wind  pressure  against 
the  floor  system  will  act  in  the  lower  chord. 

With  the  wind  pressure  between  thirty  and  forty  pounds 
per  square  foot,  the  following  loads  may  be  taken  at  the 
various  panel  points  : 


SWAV   BRA  CING.  3  7 1 

At  C,  G\  K',  L,  M\  N\  C,  G,  A',  L,  M,  N. 0.35  tons. 

"  n  sindO' 0.18     " 

"  Intermediate  points 0-35     " 

"  ^  and  (9 0.18     " 

"  Intermediate  points 3.01     " 


\  '  /^ 


Fig.  I. 


I 


Fig.  2. 


M  N 


'A 

\y 

V 

V 

/ 

\ 

ob/ 

/ 

e' 

E             \ 

/\ 

/\ 

/\ 

/ 

N°'^ 

<w'/h 

' 

J      f 

s 

r^ 

S  K  L M    N 


\  / 


M  N 


^ 


The  amount  3.01  tons  invoK-es  the  pressure  against  the 
train,  which  is  taken  at  300  pounds  per  foot  of  track.  The 
panel  length  is  fourteen  feet,  hence  the  panel  train  load  is 
14  X  300  =  4,200  pounds  =2.10  tons.  The  wind  pressure 
against  the  floor  system  is  assumed  to  be  0.56  ton  per  panel, 
while  the  panel  pressure  against  each  truss  is  0.35  ton.  The 
sum  of  the  three  quantities  is  3.01  tons. 

The  panel  train  loads  (2.10  tojis)  constitute  a  continuous,  mov- 
ing load  ;  the  wind  pressure  against  the  trusses  and  floor  sys- 
tem, however,  forms  a  fixed  load. 

The  following  are  the  truss  dimensions,  including  the 
lengths  of  the  braces  .-i/^and  A' F' : 


Panel  length  =  14.00  feet. 
Width,    BD  =  14.00     " 
CF  =     7.00     " 


Height  of  truss  =  16.00  feet. 
BC  =  21.26     " 
AC=    4-04     '* 


FA  =  8.08  feet. 


-72  WIND    S7RESSES  AND   BRACED   PIERS. 

Normal  from  C  on  FA  =  CF  x  siit  30°  =  3.5  feet. 

Let  H  represent  half  the  total  wind  pressure  concentrated 
in  the  two  upper  chords;  this  will  be  resisted  (if  the  bridge  is 
not  blown  bodily  off  the  abutments  or  piers)  by  an  equal 
force  of  friction  developed  at  the  feet  B  and  D,  or  O  and  O' 
of  the  end  posts.  Let  H'  and  H"  be  the  forces  developed  at 
D  and  B,  respectively.  These  horizontal  forces  will  tend  to 
overturn  the  trusses  in  a  vertical  plane  normal  to  the  axis  of 
the  bridge.  A  vertically  upward  reaction,  V,  will  be  devel- 
oped at  B,  and  an  equal  downward  one  (a  portion  of  the 
weight  of  the  truss  DC'N'O')  at  C.  Considering  the  left 
end  of  the  truss,  the  following  three  conditional  equations  of 
equilibrium  must  be  fulfilled  : 

H'+  H"^-  H  =  o     .     .     ,     ,     .     .     .     (i). 
F+F'=o    .......     (2). 

(//'  +  //")  X  16 +Fx  14  =  0 (3). 

The  vertical  force  acting  at  C  is  represented  by  V. 

These  three  equations  are  not  sufficient  for  the  determina- 
tion of  the  four  quantities  H',  H",  V,  and  V.  The  forces//' 
and  //"  are  therefore  indeterminate  in  magnitude,  except  in 
this  respect,  their  sum  must  be  equal  and  opposite  to  H. 

With  the  form  of  portal  bracing  shown  in  Fig.  i,  it  will  be 
assumed  in  this  article  that  H"  =  o  and  H'  —  —  H.  Other 
and  better  forms  of  portal,  together  with  other  assumptions 
in  regard  to  the  horizontal  reactions  H'  and  //",  will  be  given 
in  succeeding  articles. 

From  the  data  already  given : 

//"  =  6  X  0.35  =  2.10  tons. 

Hence  by  Eqs.  (3)  and  (2)  ; 

F=  2.40  tons  =  —  F'. 
At   the  point  E  let   there   be  supposed  to  act  two  forces 


SIVA  V  BRACIXG. 


373 


equal,  opposite,  and  parallel  to  H  and  H' ;  these  two  forces 
will  balance  each  other.  Instead  of  the  two  forces  //  and 
//,  there  may  then  be  taken  two  couples,  M'  =  H'  x  14,  and 
M  =  Hxi6. 

In  the  same  manner  at  E  let  two  forces  equal,  opposite, 
and  parallel  to  F  and  V  be  supposed  to  act.  Then,  instead 
of  Fand  V .  there  will  exist  cwo  couples,  M"  =  V  x  14,  and 
M"'  =  V  X  14. 

The  couples  whose  moments  are  Af  and  M'"  balance  each 
other,  as  is  shown  by  Eq.  (3).  The  couples  whose  moments 
are  M'and  M"  have  axes  at  right  angles,  consequently  their 
resultant  will  be : 


yj/j  =  \/M'^+  J/"2  =  14  V  (2.1)2 -I-  (2.4)3  =  44.66  ft.  tons. 

The  plane  in  which  M^  acts,  contains  the  chords  BO  and 
C'JV\  and  the  direction  of  the  couple  is  such  that  it  causes 
compression  in  C'N'  and  tension  in  BO.  Consequently  for 
those  stresses 

(BO)  =  -  (C'N')  =  44.66  ^  21.26  =+  2.1  tons. 

As  a  check  ;  (BO)  =  +   F  x    14  ^  16  =  +  2.1  tons. 

The  following  stresses  in  the  members  of  the  portal  of  the 
bridge  may  now  be  written  : 

(A'F')  —  H  X  21.26  -^  3.5  =  +  12.8  tons. 
(A'C)  =  -  12.8   X  sin  30°  =  -    6.4     " 
(FC')  =  —  12.8  X  cos  30°  =  —  ii.i     " 

The  greatest  bending  moment  in  DC  exists  at  F ,  and  is: 

M^  =  2.1   X  (21.26  —  7.00)  =  29.95  ft.  tons. 

The  stress  in  BC  is  : 

/„„,  21.26       ^_  21.26       ., 

(BC)  — X  H-= ^-x  V  =  —  3.19  tons. 

^       '  14  16 


T^jA  M^/.VD    STRESSES  AND   BRACED  PIERS. 

The  compressive  stress  in  DC ,  due  to  the  vertical  loading, 
is  relieved  by  the  same  amount. 

The  greatest  bending  moment  in  CC  exists  at  .-J,  and  has 
for  its  value  : 

J/3  =  {BC)  X  (14.00  —  4.04)  =  31.77  ft.  tons. 

With  the  wind  in  the  direction  taken,  the  brace  AF  must 
be  supposed  not  to  act  at  all.  BotJi  moments  J/^  and  M^  pro- 
duce bending  in  the  pla7ie  of  the  portal. 

The  end  post  {DC),  always  of  uniform  cross  section,  must 
be  able  to  resist  with  a  proper  safety  factor,  at  F',  the  bend- 
ing moment  M^.  The  sway  brace  CC  must  be  able  to  resist 
the  moment  M^  at  both  the  points  A  and  A' . 

The  ordinary  truss  stresses  in  the  sway  bracing  remain  to 
be  found. 

In  both  upper  and  lower  sway  bracing  the  inclined  members 
are  tension  ones  only,  while  those  normal  to  the  planes  of  the 
trusses  (in  the  direction  of  the  wind)  sustain  compression 
only. 

In  the  upper  chord  the  truss  CGMN'  has  the  two 
points  of  support  C  and  N' .  The  following  trigonometric 
quantities  will  be  required  : 

Angle  G'KK'  =  45°   ■  tan  45°  =  i. 

.yrr45"  =  1.414. 

The  upper  web  stresses  are  the  following  : 

{KK')  =  =  -  0.35  tons. 

{C K)  =  +  2  X  0.35   X  sec  Ot^"   =  +  i.oo    " 

(G'G)  =  -  3  X  0.35  =  -  1.05    " 

(GC)  =  +  4  X  0.35   X  sec  4S°   =  +  2.00    " 
(CC)  =  =  +  0.35    " 

The  resultant  upper  chord  stresses  are  the  following ; 

(CG)  =  —  4  X  0.35   X  tan  45°  =  —  1.40  tons. 

{G'M')=  -  2     "  "       "     -  1.40=  -  2.10     " 


SH^Ay  BRA  CING.  ^  -  r 

{GK)  =  +  4  X  0.35  X  tan  45°  =  +  1.40  tons. 

(AVJ  =  +  2     "         "         "     +  1.40  =  +  2.10    " 

The  lower  resultant  web  stresses  are  the  following,  remem- 
bering that  the  train*  pressure  is  a  moving  load,  and  that 
D  and  O  are  the  supporting  points  for  the  lower  sway  truss : 

{QR)=-v    6xo.30xi-rc45^  =+  2.56  tons. 

(0<2)=  -   6x0.30  -0.35  =  -  2.15 

{P'Q)  =  +  10X0.3OX  j-^<:  45°  +  2  X0.63  X5r^45''=  +  6.04 

(P/")  =  —  10x0.30  —3x0.634-    0.28  =  —  4.61 

{E'P)  =  +  15  xo.30xj-r<:45°  +  4x0.63  xj-rr 45°=  +  9.96 

(Zfi^')  =  -  15x0.30  -5x0.63+    0.28=-  7.37 

{DE)  =  +  21  X0.30X  j-rr  45°  +  6  XO.63  xsi-c  45"^=  +  14.31 

The  +  0.28  ton,  which  is  a  release,  is  due  to  the  fact  that 
the  half  panel  wind  pressure  against  the  floor  system  is 
added  to  0.35  ton,  and  taken  o?ice  too  many  times  in  each  of 
the  struts. 

The  quantity  0.30  will  be  at  once  recognized  as  2.10  -=-  7. 
The  counters  S' T  and  R'S  are  not  required  to  resist  wind 
stresses,  but  should  never  be  omitted,  in  order  that  the 
general  stiffness  of  the  bridge  may  be  increased  ;  their  cross 
sections  may  be  the  same  as  that  of  Q'R. 

The  lower  resultant  chord  stresses  are  the  following : 

{DE')  =  —(6x0.63  +  3x2.10)       tan   45°      =  —  10.08    tons. 
{E'P')=  —  (4x0.63  4-  2x2.10)        "  —  10.08  —  -  16.80      " 
(PS')  =  - (2x0.63  +        2.10)        "- 16.80  =  -20.16      " 

{EP)  =  -  {DE)  +  2.10  =  +  12.18  tons. 
{PQ)  =  -  {E'P')  +  2.10  =  +  18.90     " 
(^/e)  1=  -  (/^'S')  +  2.10  =+ 22.26     " 

Although  not  a  part  of  the  lower  chord  of  the  truss  under 
consideration,  BE  sustains  the  stress  : 

{BE)  =  +  2.10  tons. 


The  train  is  taken  as  passing  from  right  to  left. 


376  IVIND    STRESSES  AND   BRACED  PIERS. 

If  the  wind  blows  in  the  opposite  direction  to  that  assumed, 
the  chord  stresses  which  have  been  determined  for  C N'  will 
be  found  in  CN,  and  vice  versa.  Precisely  corresponding 
changes  are  to  be  made  in  the  lower  chords.  The  stresses  in 
the  sway  struts  would  not  be  changed.  That  diagonal  in 
each  panel  which  is  not  stressed  in  the  preceding  instance, 
would  sustain  a  tensile  stress  exactly  equal  to  that  already 
found  in  the  other  diagonal. 

It  is  therefore  necessary  to  make  calculations  for  but  one 
direction  of  the  wind. 

So  far  as  equilibrium  is  concerned,  in  the  preceding  inves- 
tigation, there  might  be  taken  H"  =  —  //  and  H'  =  o.  In 
such  a  case  ^C  would  be  subjected  to  a  bending  moment  at 
F  equal  to  —  jW.^  ;  and  the  bending  moment  in  CC,  at  A, 
would  be  —  J/3;  while  the  stresses  in  FA,  AC,  and  CF  would 
be  respectively  —  {FA'),  —  (A'C),  and  —  (C'F').  For  these 
reasons  all  parts  of  the  portal  should  be  built  to  sustain  the 
stresses  and  moments  which  have  been  found  when  affected 
by  opposite  signs. 

It  should  be  remembered  that  the  parts  EC,  E'C,  and  CC 
are  subjected  to  combined  direct  stresses  and  bendings  to 
the  respective  amounts  that  have  been  found. 

Those  portions  of  the  lower  sway  struts  £E',  PF ,  etc., 
extending  from  the  windward  rail  to  the  lower  chord  BO 
(with  the  direction  of  the  wind  first  assumed),  are  each 
subjected  to  a  compressive  stress,  in  addition  to  those  already 
found,  nearly  equal  to  an  amount  to  be  determined  in  the 
following  manner  :  Let  N  represent  the  number  of  panels 
in  the  sway  truss,  and  n  the  number  of  any  strut,  from  the 
farther  end  of  the  truss,  counting  the  end  itself  zero,  i.  e.,  for 
PP' ,  n  will  be  5.  In  the  case  taken  ^V  =  7.  The  amount 
desired  will  then  be  the  panel  train  wind  load  multiplied  byJL 

isr 

added  to  the  panel  wind  pressure  against  the  floor  system  ;  or  in 
the  example, 

2.10  X 1-   0.56  =  0.30  X   n  +  0.56. 


SIVAY   BRA  CING.  UJ 

This  compression  in  the  struts  arises  from  the  fact  that  the 
wind  pressure  against  train  and  floor  system  is  not  applied 
at  panel  points,  but  on  the  struts  betzveen  their  ends,  and  that 
a  panel  load  of  the  former  must  be  added  to  the  ordinary- 
strut  stress  which  exists  with  the  head  of  the  train  at  the 
strut  considered. 

This  involves,  however,  a  very  small  error  on  the  side  of 
safety,  since  the  pressure  is  divided  between  the  two  rails. 
Considering  both  directions  of  the  wind,  it  will  be  seen  that 
these  struts  are  subjected  to  this  amount  of  compression 
from  end  to  end,  in  addition  to  the  regular  truss  stresses. 

All  the  preceding  wind  stresses  are  to  be  combined  with 
those  due  to  the  vertical  loading,  wherever  they  act  in  the 
same  piece. 

If  the  portals  are  vertical,  the  stresses  {BO)  and  (O'N'),  due 
to  the  moment  J/j,  will  be  zero  ;  also  the  span  of  the  upper 
sway  truss  will  be  equal  to  that  of  the  lower.  No  other 
changes  will  occur. 

If  the  bridge  is  a  deck  one,  when  possible,  the  ends  of  the 
chords  should  be  secured  directly  to  the  piers  or  abutments, 
as  no  bending  will  then  take  place  in  the  end  posts.  If  this 
is  not  possible,  the  calculations  will  be  precisely  the  same  as 
those  already  indicated,  with  possible  changes  of  signs  in 
some  of  the  stresses  in  the  end  posts  or  braces.  In  deck 
bridges,  however,  the  wind  pressure  against  floor  system  and 
train  will  be  found  in  the  upper  chord. 

The  method  of  treatment  which  has  been  exemplified  is, 
therefore,  perfectly  general  and  sufficient  for  all  cases. 

In  deck  bridges,  tension  sway  braces  (contained  in  planes 
normal  to  the  trusses)  are  introduced,  extending  from  either 
chord  of  one  truss  to  the  diagonally  opposite  one  in  the 
adjacent  truss.  So  far  as  pure  equilibrium  is  concerned,  when 
horizontal  sway  trusses  are  present,  these  are  superfluous  ;  but 
they  are  very  efficient  in  their  influence  on  lateral  stability. 
The  actual  stresses,  in  these  members,  in  any  given  case,  are 
indeterminate,  but  their  greatest  possible  values  are  easily 
fixed.  Let  the  total  wind  pressure  exerted  at  a  pair  of  oppo- 
site  panel    points  in  the  two  upper  chords  be  represented  by 


378  WIND    STRESSES  AND   BRACED   PIERS. 

P ,  and  let  a  represent  the  angle  between  a  horizontal  line 
and  the  tension  brace  in  question.  Then  the  greatest  possible 
stress  which  is  required  will  be  :    T'  —  P  sec  n. 

This  assumes  that  all  the  wind  pressure  is  carried  to  the 
lower  chords  and  resisted  by  the  lower  sway  truss. 

In  the  case  of  vertical  end  posts,  where  the  upper  chord 
ends  are  not  secured  directly  to  piers  or  abutments,  the 
stresses  in  the  end  lateral  diagonals  become  perfectly  deter- 
minate. In  fact,  in  such  a  case,  the  braces  AF,  A'F',  Figs. 
I  and  2,  become  the  diagonals  in  question.  Let  P  represent 
half  the  total  wind  pressure  in  the  two  upper  chords.  The 
tensile  stress  in  either  one  of  these  diagonals  (if  a  retains  its 
preceding  signification)  will  then  be : 

T  ^  Px  sec  a. 

Let  P  represent  the  total  wind  pressure  against  the  bridge, 
and  P"  the  total  wind  pressure  against  the  train  when  it  covers 
the  whole  span.  Then  let  W  and  W  represent  the  total 
weight  of  bridge  and  train  respectively  ;  also  let  /  be  the 
coefficient  of  friction  between  the  foot  of  end  post  and  the 
supporting  surface  underneath.  In  order  that  neither  truss 
shall  possibly  be  moved  bodily  by  the  wind,  with  the  bridge 
empty  or  covered  by  a  train,  there  must  exist  the  following 
relations : 

^  <  ^(|'_  .k)  ;  or, />./>"</ ('*:±i?:- .F.). 

In  the  cases  of  through  bridges,  or  of  deck  bridges  with 
upper  chords  not  secured  to  abutments,  Fj  is  to  be  found  by 
applying  the  general  form  of  Eq.  (3)  to  both  bridges  and  train. 

If  the  ends  of  the  chords  are  secured  to  the  piers  or  abut- 
ments, the  resistances  of  these  fastenings  will  take  the  place 
of  the  frictional  resistances. 

One  other  effect  of  the  wind  pressure  against  the  train 
remains  to  be  noticed.  The  normal  action  of  this  pressure 
will  not  permit  the  train's  weight  to  be  distributed  between 
the   two   chords   which    carry  it,  according   to  the  law  of  the 


SWAY  BRACING. 


379 


lever.  Let  /"  represent  a  panel  wind  pressure  against  the 
train,  and  let  h  represent  the  height  of  its  centre  of  ac- 
tion above  the  points  of  support.  Also  let  b  represent  the 
horizontal  distance  between  centres  of  trusses;  then 


t  =  f:A 


i        (4), 


will  be  the  amount  of  load  which  is  transferred  from  the 
windward  to  the  leeward  truss.  In  other  words,  the  panel 
leeward  load  will  exceed  the  panel  windward  one  by  2t.  If, 
therefore,  lu'  is  a  panel  moving  load,  the  action  of  the  wind 
will  cause  all  moving  load  truss  stresses  to  be  increased  by  an 
amount  found  by  multiplying  the  stresses,  determined  with- 
out regard  to  the  wind,  by —  .     Also,    if   s    is    the     distance 

(normal)  between  two  adjacent  parallel  stringers,  the  increase 
of  load  on  one,  and  decrease  of  that  on  the  other,  will  be: 

t,  =-—7-^ (5). 


Eq.  (5)  gives  the  variation  of  load  on  the  floor  beam  also. 
Without  essential  error,  h  may  be  measured  downward  from 
the  centre  of  the  body  of  the  car. 

Specifications  sometimes  require  calculations  to  be  made 
with  an  unloaded  bridge.  In  such  a  case  the  methods  are 
precisely  the  same  as  the  preceding,  with  the  train  wind 
pressure  omitted. 


Art.  82. — Transverse  Bracing  for  Transferring  Wind   Stresses   from  One 
Chord  to  Another — Concentrated   Reaction. 

In  the  preceding  Article  it  has  been  supposed  that  the  wind 
pressure  is  resisted  by  sway  trusses  in  the  horizontal  planes  of 
both  upper  and  lower  chord.  It  may  sometimes  be  desirable 
to  transfer  all  the  wind  pressure  to  the  lower  chord,  or  to 
the  upper. 


38o 


WIND    STRESSES  AND  BRACED   PIERS. 


Ul-UI 
itidm 


Fig.  I. 


The    section    of   a   through    truss    bridge,   in    which    it  is 
desired  to  carry  all  the  wind   pressure  to  the  lower  chord,  is 
represented  in   Fig.  i.     AC  and  ND  are 
posts  directly   opposite    to  each   other  in 
the  two  trusses.     AN  and  OB  are  lateral 
struts,  while  AO  and  BN  are  lateral   ties. 
Let  the  wind  be  supposed  to   blow  from 
right    to    left,    as    shown    by    the     arrow. 
According    to   the   principles  of  the  pre- 
ceding   Article,    in     consequence    of    its 
direction  the   wind  will  relieve  the  truss 
AC   of    a    part    of    the    weight    which   it 
carries,   and    add   the   same    amount  to  that   carried   by  the 
truss  BN. 

If  the  direction  of  the  wind  were  reversed,  the  truss  DN 
would  be  relieved,  and  AC  would  receive  the  increase  of 
loading. 

Let  this  relief  (or  increase)  of  truss  load,  per  panel,  be  de- 
noted by  w  ;  it  zvill  act  as  tJiough  hung  from  B. 
The  following  notation,  also,  will  be  used  : 

AB  =  d  ^  ON.  BC=  a  ^  OD. 

DC  =  AN  =  b. 

F  =  total   wind    pressure,    per  panel    (for   one  truss),   on 

U^B  +  BC). 
F'   =   total   wind   pressure,  per    panel    (for  one  truss),  on 

lAB. 


With  the  assumed  direction  of  the  wind,  the  tie  ^6^  will 
not  be  stressed.  As  usual,  the  plus  sign  (  +  )  will  indicate 
tension,  while  the  minus  (— )  sign  will  indicate  compression. 

In  this  article  the  total  horizontal  reaction,  equal  to 
2{F  +  F'),  will  be  taken  as  concentrated  at  D. 


TRANSVERSE   BRACIA'i7.  381 

There  will  then  result : 

Relief  in  truss /J C:  =  w  =  2f — ^^ j^ J.    .    (i). 

Compression  in  JzV=        —  {AN)  =  —  F'.      .     .     (2), 

Tension  in  BlV         =        +  {BN)  =  +  wsecABN.  (3). 

Compression  in  5(9  =       -  {BO)  -  -  {F  +  w  tan  ABN) 

2a^ 


=       -  (F  +  2F'  +  [F  +  F']  ^^ 
=  F-2iF^F')   (^^).    (4). 

Compression  in  ND  —       —  {^'D)  =  —  w.     .      .     (5). 

The  horizontal  force  2{F -^  F')  acts  toward  C,  at  Z>,  pro- 
ducing a  bending  in  DN  which,  has  its  greatest  moment  M 
at  O.      Hence : 

M=2{F-hF')a (6). 

If  A' is  the  greatest  intensity  of  compressive  stress  (due  tc 
flexure)  in  the  cross  section  of  the  post  DN,  at  O,  d^  the 
greatest  distance  of  any  compressed  fibre  from  the  neutral 
axis  of  the  cross  section,  and  /the  moment  of  inertia  of  the 
section  about  an  axis  passing  through  its  centre  of  gravity 
and  lying  in  the  plane  of  the  truss  ;  then,  by  the  well-known 
formula — 

K  =  '!^ (7). 

At  O  there  will  then  exist  the  intensity  of  compression  : 

—  { —  +  A^)  ;  in  which  q  is  the  area  of  cross  section  of 


382  WIND   STRESSES  AND    bRACED   PIERS 

the  column.      The    intensity   of  compression  —  (  —  -I-  K]    is- 

\^  / 

in  addition  to  the  regular  truss  stresses  arising   from  vertical 
and  wind  loads. 

If,  for  lack  of  head  room,  a  flanged  beam  only  is  used, 
as  shown  in  Fig.  2,  instead  of  the  lateral  bracing  of  Fig, 
I,  then  that  beam  will  be  subjected  to  combined  compres- 
sion and  bending.  Let  +  7^  represent  the  total  wind  pressure, 
per  panel,  for  both  trusses,  on  \A  />,  and  let  w 
represent  the  release  of  weight  in  AB  and  in- 
crease in  CD. 
Also  let 

AB  =  a,  and  BD  —  b.      \F  is  to  be  taken  as 
dJMI  M^  applied  at  yJ  and  zv,  at  the  same  point.     Equal 

and  opposite  forces  are  also  to  be  supposed 
to  act  at  D.     The  moment 

M  =  Fa  =  ivb,  exists  at  C  and  gives: 


BA 


Fa 

With  the  direction  of  wind  shown  by  the  arrow,  the  bend- 
ing caused  by  w  will  increase  uniformly  from  nothing  at  A  to 

M  =  ivb 

at  C.  The  bending  moment,  therefore,  to  be  resisted  by  this 
beam  A  C,  and  by  the  Joints  between  it  and  the  chords  A  and  C, 
is : 

M=Fa  =  wb (8). 

The  direct  compression  in  AC  is: 

(AC)=  -^F. (9). 

Hence,  if  q  is  the  area  of  cross  section  of  the  beam,  and  if 
K,  /,  and  di  retain  the  same  general  signification  as  in  Eq.  (7), 
the  greatest  intensity  of  compression  in  the  beam  (at  its 
ends)  will  be : 


TJiANSVERSE   BRACING.  g 

-(-;h-^^ (-)• 

The  direct  compression  in  CD  is 

{CD)  =^  -w (II). 

The  bending  moment  in  CD  at  C  is : 

M=Fa (12). 

The  greatest   compressive  intensity  is   found  at  once  by 
Eq.  (10),  after  writing  2U  for  IF,  and   giving  to 
the  remaining  notation  its  general  signification. 

The  two  preceding  cases  are  those  of  through 
trusses.  In  the  case  of  a  deck  truss  the  lateral 
bracing  is  of  much  more  simple  character  ;  it  is 
shown  in  Fig.  3.  At  C  and  A  are  the  two  lower 
chords.  CA  is  a  lateral  strut,  while  BC  and  cj|l 
DA  are  lateral  ties.  No  parts  are  subjected  to 
bending. 

If  F  is  the  panel  wind  pressure  (for  both  trusses)  acting 
along  AC,  there  will  result  : 


Fig. 


{CA)  =  -  IF 
{BA)  =  -  zv. 


(13)- 
(14). 


{BC)  =  +  VF^  +  W' (15). 

The  horizontal  component  of  {BC)  is  equal  to  F,  and 
acts  at  B.  Thus  all  wind  pressure  is  carried  to  the  upper 
chord. 

The  compression  {BA)  is  in  addition  to  the  regular  truss 
stresses  induced  by  the  vertical  and  wind  loads. 

By  these  methods  all  the  wind  pressure  may  be  carried  to 
either  chord.  The  truss  stresses  of  the  sway  truss  in  the 
horizontal  plane  of  that  chord  have  already  been  found  in 
the     preceding  Article,  or    rather,  the   methods   for   finding 


384 


WIND    STRESSES  AND   BRACED   PIERS. 


them  and   the   effect   of  iv  on   the   stresses   in    the    vertical 
trusses  have  there  been  completely  given. 

The  wind  has  been  taken  in  one  direction  only;  with  the 
other  direction,  opposite  but  symmetrically  located  parts 
would  be  stressed  by  the  amounts  found. 

Art.  83. — Transverse  Bracing   with   Distributed  Reactions. 

In  the  preceding  articles  it  has  been  assumed  that  the  hori- 
zontal reactions  of  the  wind  pressure  were  concentrated  at  the 
extremity  (top  or  bottom,  as  the  case  may  be)  of  one  post  in 
the  tranve^se  panel  considered.  This  assumption,  however 
may  not  be  admitted ;  or  some  other  may  be  substituted  in 
its  place. 

Let  Fig.  I  represent  a  transverse  panel,  with  the  wind 
blowing  in  the  direction  shown  by  the  arrow; 

As  before,  the  following  notation  will  be  used  : 

AB  =  ON  ^  d.  BC=  OD  =  a. 

DC  =  AN  =  b. 


F  =total  wind  pressure,  per  panel,  for  one  truss,  on  \{AB-\-  BC). 

rp' <t  u  it  n  <i  u  a 

w  =  relief  of  load  in  truss  AC. 


hAB. 


Instead  of  concentrating  the  entire  hori- 
zontal reaction  at  D,  if  «  is  a  quantity  less 
than  unity,  there  will  be  assumed: 

Horizontal  reaction  at  Z)  =  2n{F  +  F). 

"     C  =  2{i  - n){F  +  F'). 

The   wind    pressures  on  ^BC  =  ^OD  act 
directly  at  C  and  D  in  the  horizontal  sway 
truss,   and,   consequently,  will  be  omitted 
from  consideration. 

As  in  the  preceding  article  : 


111    Dl 
IimID 


Hon 


Fig.   I. 


W 


2{F  {a  +  d)  +Fa) 


.     .  (I). 


TRA  NS  VERSE  BRA  CING. 
Taking  moments  about  ^: 

{AN)  =  - 


3«5 


'F'+2{i-n)(F  +  F')^'^,     .     .    (2). 
Taking  moments  about  N : 

(^^)=-[^^+^'H^  +  ^)-/r)1      .     .     .      (3). 

Taking  moments  about  the  intersection  of  AN  a.nd  OB  at 
the  distance  infinity  ( co)  from  the  figure : 

(^A^)oo  cos  ABN=-^[zvca  +  2n{F  +  F')a]  ; 
•.    {BN)  =+w  sec  ABN  =  ^l^J^l±jl±l^-secABN.  (4). 

The  stress  in  BN,  therefore,  remains  the  same  whatever 
may  be  the  assumptions  in  regard  to  the  horizontal  reactions. 

The  bending  moment  at  6>,  about  an  axis  lying  in  the  plane 
of  the  vertical  truss,  will  be  : 

M  =  2fi{F  +F')a (5). 

Since  the  windward  truss  is  always  relieved  of  a  part  of  its 
weight  the  bending  moment  2{i  —  n)  {F  +  F')a,  at  ^,  will 
seldom  or  never  be  needed. 

The  value  of  M,  from  Eq.  (5),  put  in  Eq.  (7)  of  the  pre- 
ceding Article,  and  in  the  expression  following  that  equation, 
will  enable  the  greatest  compressive  intensity  in  the  post  to 
be  found. 

If  the  transverse  panel,  Fig.  i,  represents  the  portal  of  a 
bridge,  the  distances  AB  and  BC,  or  ^^  and  a,  represent  tn- 
clincd  distances  in  the  plane  of  the  portal.  F '  will  {or  may) 
then  include,  also,  the  reaction  of  the  horizontal  sway  truss  in 
the  plane  of  AN,  while  F  will  include  the  reaction  of  the 
horizontal  sway  truss  in  the  plane  of  OB,  if  there  is  such  a 
sway  truss. 

95 


$86  WIND   STRESSES  AND  BRACED  PIERS, 

If  «  =  I,  as  is  sometimes  assumed  : 

{AN)=-\j' ^{F  ■\- F')^ (6). 

i^OB)  =  -  1"/"+  (F  +  F')--!  =  (AN).  .     .     .  (7). 

If  «  =  I  in  the  formulae  of  this  Article,  those  of  the  cor 
responding  cases  in  the  preceding  Article  at  once  follow. 
In  Fig.  2  let  the  notation  be  as  follows: 

AB=CD^  a.  BD=  CA  =  b. 

Total  wind  pressure  for  both  trusses,  per  panel,  along 

AC=F. 
Horizontal  reaction  at  Z?  =  nF. (8), 


B={i-n)F. 


(9). 


If  Fig,  2  represents  a  portal,  F  will   or  may  include  the 
reaction  of  a  horizontal  sway  truss. 

The  bending  moment  on  both  DC  and  CA, 
S*    at  C,  also  on  the  joint  at  the  same  point,  is  : 


m 


Ml  =  nFa. 


(lO). 


KiG. 


fJjjB       This  is  the  greatest  bending  in  DC  and  CA 
of  that  kind  which  produces  compression  in  the 
lower  flange  of  the  beam  CA. 
The  relief  of  panel  load  in  the  truss  AB  and  increase  of 
that  in  CD  is  ; 


w 


Fa_ 
b 


(10). 


Let  X  represent    any    variable    portion  of  CA ;    then    the 
bending  moment  at  any  point  of  CA  is: 


ST/(£SS£S  IN  BRACED   PIERS. 


387 


Fa 
M  =  n  Fa  —  wx  =  n  Fa 7-  x.      .     .    (i  i). 

For  the  point  or  joint  A,  x  becomes  equal  to  b,  while  the 
expression  for  the  moment  is  : 

M\^  ~  {\  -  n)  Fa (12). 

This  is  the  greatest  bending  of  the  kind  opposite  to  M ^,  in 
CA.  It  is  also  the  greatest  bending  in  AB.  The  con- 
nections at  C  must  resist  the  moment  M-^,  while  those  at  A 
must  resist  M\. 

The  direct  compression  in  CA  is  4/^ 

"   CD    "    w. 

\{n  =  \\    M,^-M\  =  \Fa.     ....     (13). 

These  various  bending  moments,  substituted  in  Eq.  (10)  of 
the  preceding  Article  for  J/,  will  enable  the  greatest  intensi- 
ties of  stress  in  the  members  CA,  CD,  and  y^^  to  be  at  once 
found. 

Art.  84. — Stresses  in  Braced  Piers. 

The  general  treatment  of  stresses  in  braced  piers  may  be 
exemplified  by  that  of  a  single  "  bent  "  represented  by  a 
skeleton  diagram  in  Fig.  i,  in  which  the  horizonal  web  mem- 
bers are  compressive  ones.  The  plane  of  the  "  bent  "  is  verti- 
cal and  normal  to  the  centre  line  of  the  truss  whose  end  rests 
upon  it ;  or  if  the  track  is  curved,  this  plane  is  normal  to  it. 
The  bent  shown  in  Fig.  i  may  be  considered  one  of  a  pair,  in 
parallel  planes,  which,  being  braced  together,  compose  the 
complete  braced  pier.  The  dotted  rectangle  .iJ/A^.5  repre- 
sents a  skeleton  section  of  the  truss  supported  by  the  piers, 
the  upper  chords  of  which  rest  upon  the  top  of  the  pier  at  A 
and  B.     A  skeleton  section  of  the  train  is  also  shown. 

The  direction  of  the  wind  is  supposed  to  be  shown  by  the 
arrows  a,  normal  to  the  track  at  the  top  of  the  pier.     If  the 


388 


WIND    STRESSES  AND  BRACED  PIERS. 


trusses  are  loaded  with   a  train,  the  wind  pressure  against 
them  and  the  train  will  be  carried  to  the  top  of  the  piers  in 

the  manner  shown  in  Art,  8i.  The 
wind  will  also  act  against  the  pier 
itself. 

Let  the  train  be  supposed  to  cover 
the  whole  of  the  two  spans  adjacent 
to  the  top  of  the  bent  (in  all  ordi- 
nary cases  one  of  these  spans  will  be 
the  distance  between  two  adjacent 
bents) ;  then  let  H  represent  half 
the  total  pressure  against  trusses, 
and  P{  half  that  on  the  train  cover- 
ing the  two  spans. 

The  height  of  the  centre  of  action 
of  /^i"  above  AB,  Fig.  i,  is  h.  Also 
let  b  =  AB.  The  pressure  /\"  will 
decrease  the  train  reaction  at  A  and 

increase  that  at  B  by  the  amount : 
Fig.  1.  ^ 


K.=  ^'. 


(I) 


Let  k'  represent  the  vertical  distance  of  the  centre  of  ac- 
tion of  H  from  the  horizontal  line  AB. 

The  wind  pressure  on  the  truss  AMNB  will  cause  an  in- 
crease of  truss  reaction  at  A,  and  an  equal  decrease  of  that 
at  B,  which  will  be  denoted  by  F,  and  its  value  will  be: 


V  = 


Hh' 


consequently  if 


/'=  Fi-  F  = 


P;'h 


Hh' 

b 


the  total  horizofital  force  to  be  taken  as  acting  at  A,  and  with 
the  wind,  will  be  (//  +  /\"  —  2/'  tan  a)  added  to  the  wind 
pressure  acting  directly  at  A.     In  Fig.  2,  r^  represents  this 


STJ?£SS£S  IX  BRACED   PIERS. 


;89 


force,  laid  down  to  any  desired  scale, 
measured  to  the  right  of  d  represent 
the  panel  wind  pressures  against  the 
pier  at  the  points  C,  E,  G,  and  A', 
while  those  shown  on  the  left  of  c  rep- 
resent the  panel  pressures  at  D,  B, 
F,  H,  and  L.  The  panel  pressures  at 
A,  B,  K,  and  L  are  half  those  at  the 
other  points. 

Let  W  represent  the  total  weight 
of  adjacent  trusses  and  moving  load 
resting  at  the  top  of  the  pier. 

Let  Wi  represent  the  panel  weight 
of  the  pier  itself  resting  at  the  points 
C,  E,  G,  D,  F,  H:  4  n\  will  be  taken 
as  applied  at  the  points  A,  B,  K,  and 
L  ;  then  the  resultant  reactions  at  A 
and  B,  with  the  wind  blowing,  will 
be,  respectively, 

—  —  t  and V  t . 

2  2 


The  small  segments 


Fig.  2. 


(2). 


It  has  been  implicitly  supposed  that  two  equal  and  opposite 
forces,  equal  in  magnitude  and  parallel  to  /\'',  act  along  AB. 
One  of  these  forms,  with  Px'  itself,  the  couple /*l'7^ ;  the  other 
is  the  wind  pressure  which,  combined  with  the  half  panel  press- 
ure at  A,  and  {H—2t'  tan  a),  is  represented  by  cd  in  Fig.  2. 

The  quantity  /'  is  the  force  of  a  couple  whose  lever  arm  is 
b.  One  force  /'  is  therefore  supposed  to  act  at  A,  and  the 
other  at  B.  Vi  will  be  considered  larger  than  V;  hence  /' 
will  act  upward  at  A  and  downward  at  B.  If  a  is  the  angle 
between  ^^or  BL  and  a  vertical  line,  the  /'  at  B  will  cause 
a  compression  m  AB  equal  to  /'  tan  «',  while  the  /'  at  A  will 
pull  to  the  left  by  the  same  amount.  Consequently  the  force 
2/'  tan  (n  will  act  on  the  point  A  and  toward  the  left. 

In  the  diagrams  and  in  the  equations  which  follow,  positive 
and  negative  signs  indicate  tensile  and  compressive  stresses, 
respectively. 


390  WIND   STRESSES  AND  BRACED  PIERS. 

The  stresses  due  to  vertical   loads   at  A    and   B,  and  the 
other  panel  points,  will  be  the  following : 

fW  W\ 

iAC)"=-[-j~r+^)seca. 


{EGj'~-[    "       -  + 


2 


{GK)'^-(  -       "  +  '^-^)     " 


2    / 


(-)--(-. ^■.'^0 

(^^)"=-("         "+-^ 


(FH)"  =  -(    "       "  +5i^ 


2 


(i7Z)"  =  -(  "     -  +  Zi^ 


{AB)"=-[^  +-^)  tana. 

(CD)"  =  -IV  ta7i  a. 
{EF)"=  -   " 
(GH)'. 


' it       «( 


{KL)"=+{^~-t  +L^ytaKa. 


The  difference  between  the  horizontal  component  in  HL 
and  {KL)    is  2t'  tan  ex,  and  it  acts  towards  the  right. 

The  stresses  caused  by  the  horizontal  wind  pressure  acting 
through  A,  /?,  Cy  D,  etc.,  are  shown  in  Fig.  2,  as  has  already 
oeen  noticed.  The  diagonals  sloping  similarly  to  AD  are  as- 
sumed not  to  be  stressed.      The  diagram  explains  itself. 

The  resultant  stresses,  finally,  are  to  be  found  by  combin- 


ST/?ESSES  IN  BRACED   PIERS. 


391 


{~DF)  = 

-(•• 

,. ,  3  ^^) 

<( 

{'FH)  = 

-(" 

.. ,  ^^ 

" 

{'Hi)  = 

-(" 

-^^) 

(( 

ing  the  results  of  the  diagram  in  Fig.  2  with  those  expressed 
by  the  equations  already  written.     They  are  the  following: 

iAC)  =  -[^-^  +  ^)-^'^-^-- 

(C£)  =  -("       "+-2-)     "      +(6^^)- 

(EG)   =  -("       "  +^!")     "     +{£G). 

(GK)  =  -("       "  +^2^)     "     +{GK). 

nv  jv\ 

(BD)  =  _  (^^  +  /'  +  -  '  j    ^^f  «  -  (BD). 

-  [DF). 

-  (FH). 

-  iHL). 

(AB)   =  -i{lV+  IV)  tan  a  -  {AB). 

(CD)  =  -  VV  tan  n  -  (CD). 
(Rf)  =  -  "  "  -  (EF). 
(GH)  =  -  "        "     -  (GN). 

(KL)  =  +  (t  -  ^'  +  -f)  '^''  «  -  (^^)- 

It  is  not  necessary  to  reproduce  the  stresses  in  the  oblique 
web  members,  since  they  can  be  scaled  directly  from  Fig.  2. 

All  the  stresses  may  be  checked  by  the  method  of 
moments  in  the'  usual  manner,  and  such  checks  should  always 
be  applied. 

The  two  reactions  R  and  R'  are  the  following  : 

R  =.  i{W  -  2t'  ^  S  IV,)  +  Ri. 
R'  =  \{W  +  2/'  +  8  \\\)  +  /?i'. 


392 


WIND    STRESSES  AND   BRACED  PIERS. 


It  is  to  be  remembered  that  R(  is  to  be  taken  as  positivt 
in  these  expressions  ;  also  that  R-l—  —  R\i  as  shown  in  Fig.  2. 

The  lateral  force  Fi  to  be  resisted  at  the  foot  of  the  bent 
by  friction  or  some  special  device,  is  the  total  wind  pressure 
against  the  train,  truss,  and  bent. 

If  y  is  the  coef^cient  of  friction  at  A' and  L,  Fig.  i,  the 
lateral  resistance  of  friction  offered  at  K  is  f'R,  and  that 
at  L,f'R'.  It  is  supposed  that  both  the  reactions  R  and 
R'  are  upward,  also  that  both  coefficients  of  friction  are  the 
same. 

The  expression  for  {KL)  has  been  written  on  the  assump- 
tion that  all  frictional  resistance  is  exerted  at  L.  Strictly, 
however,  the  stress  in  KL  may  be  taken  as : 

{KL\  =  {'kL)  -f'R;      . 

always  supposing,  numerically,  {KL)  >  /R. 

The  circumstances  of  particular  cases  frequently  require 
calculations  to  be  made  with  the  structure  free  of  moving 
load,  as  well  as  covered  with  it.  In  such  a  case  it  is  only 
necessary  to  put  for  IV,  in  the  preceding  operations,  the 
weight  of  trusses  only. 

The  wind  has  been  taken  in  but  one  direction  only,  though 
the  pier  is  to  be  designed  for  both  directions,  since  it  is  only 
necessary  in  the  resultant  stresses  to  change  the  letters  B,  D, 
F,  LL,  L,  to  A,  C,  E,  G,  K,  and  vice  versa. 

If  MN,  Fig.  I,  should  coincide  with  AB,  or  if  the  truss 
should  rest  upon  the  top  of  the  pier,  it  would  only  be 
necessary  to  take  t'  =  Fj  +  V,  remembering  that  h  is  the 
distance  (vertical)  from  the  centre  of  /-*/'  to  the  top  of  the 
pier. 

It  should  be  stated  that  2t'  tan  a  may  be  treated  as  a 
single  force  acting  toward  the  left  and  along  AB,  Fig.  i.  It 
will  then  give  rise  to  the  diagram  in  Fig.  3,  which  shows  all 
the  stresses  produced  by  its  action.  In  that  Fig.  ad  repre- 
sents   2/'  tan  a.      In  such   a  treatment   of  the  question,   cd. 


STRESSES  IN  BRACED   PIERS. 


393 


Fig.  2,  would   represent  H  +  A"  added   to    the  half  panel 

pressure  at  A.     The  resultant  stresses  would  then  be  found 

by  combining  the  results  of  the  two  diagrams  with  those  of 

the  equations.     All  the  results  of  these 

two  methods  will  not  agree  ;  the  latter 

will  give   the  greatest.     This  ambiguity 

cannot   be   av^oided,  for   it    results   from 

the    fact    that    the    pier    cannot    be    so 

divided    as    to     sever    these     members 

only. 

Mr.  J.  A.  Powers,  C.  E.,  has  called  the 
attention  of  the  writer  to  the  fact  that 
the  web  members  of  a  braced  pier  carry- 
ing a  double  track  railway,  similar  to 
that  shown  in  Fig.  4,  will  receive  their 
greatest  stresses  with  the  windward  track 
only  loaded. 

The  vertical  member  GH  vazy  be  sup- 
posed to  carry  its  proper  proportion  of 
the  load  which  rests  on  each  track.    This  ^^^-  3- 

supposition,  however,    does  not    affect  the  statement  made 

above. 

Let  the  wind  have  the  direc- 
tion shown  by  the  arrow,  and  let 
\\\  as  before,  represent  the  fixed 
and  moving  weight  resting  on 
GB,  while  iv  is  that  part  of  W 
which  is  carried  to  B.  If  GH 
acts  • 

QD 


Fig  4. 


If  GH  does  not  act : 


CF 


W. 


DF 

In  the  latter  case,  the  beam  AB  will  carry  Wi'=  ^  W  to  A. 


394  IVJND    STRESSES  AND   BRACED   PIERS. 

If  the  angle  FBN  =  CAM—  a,  the  force 

h  =  zv'  tan  a  —  u>i  tan  a 

will  act  along  /^i5  as  an  unbalanced  horizontal  one.  If  GH 
acts,  Wx  tan  a  becomes  equal  to  zero. 

Then  in  the  preceding  investigation,  there  is  to  be  put, 
(//  +  Ji)  for  //,  while  iv'  is  to  be  taken  as  acting  vertically- 
down  at  B,  and  zi\  or  o  (as  the  case  may  be)  at  A.  The 
preceding  methods  and  diagrams  remain  exactly  the  same  as 
before. 

In   the    formulae,   however,  w/  or  o  is  to  be  put  for   the  

2 

at  A,  Fig.  I,  and  w'  for  that  at  B  in  the  same  figure.  Noth- 
ing else  is  changed. 

If  W  rests  on  AG  and  GB  at  the  same  time,  a  horizontal 
force  equal  and  opposite  to  h  is  developed  at  A^  Fig.  4. 
Hence  Ji  will  be  balanced  and  disappear. 

If  W  rests  on  AG  alone,  with  the  direction  of  the  wind 
remaijiing  the  same,  Ji  will  change  its  direction,  thus  giving 
much  smaller  web  stresses  than  those  existing  with  W  on 
GB  alone. 

If,  for  any  reason,  the  load  on  a  single  track  pier  does  not 
rest  over  its  centre,  h  will  have  a  definite  value,  and  the  above 
considerations  must  govern  the  determination  of  the  web 
stresses.  This  condition  may  exist  if  it  becomes  necessary 
to  place  braced  piers  under  a  single  track  railway  curve. 

The  stresses  caused  by  the  traction,  or  pull,  of  the  loco- 
motive, in  the  members  of  a  braced  pier,  are  simple  in  char- 
acter and  easily  determined. 

In  such  a  case,  the  pier  is  simply  a  cantilever  with  the 
traction,  or  pull,  as  a  single  force  acting  at  its  extremity. 
The  traction  acts  along  the  line  of  the  rails,  and  the  length 
of  the  cantilever  is  the  height  of  the  pier.  The  stresses  thus 
determined  are  to  be  combined  with  those  already  found. 


COMPLETE   DESIGN  OF  A    RAILWAY  BRIDGE.  395 

Art.  85. — Complete  Design  of  a  Railway  Bridge. 

The  main  sections  and  details  of  this  design  shown  on  Pis. 
XI.  and  XII.  arc  based  on  the  following  general  specifica- 
tions. 

The  span  length,  depth  of  truss,  panel  division,  moving 
load,  fixed  loads,  and.  stresses  resulting  from  the  preceding, 
shall  be  as  determined  in  Art.  ii. 

The  clear  width  between  trusses  shall  be  14  feet. 

A  wind  load  of  150  pounds  per  lin.  ft.  of  span  shall  be 
taken  for  the  upper  chord,  and  the  same  amount  for  the 
lower,  and  shall  be  treated  as  a  fixed  load  in  each  chord.  In 
addition  to  this  fixed  load,  300  pounds  per  lin.  ft.  of  span 
shall  be  taken  as  a  moving  wind  load  for  the  lower  chord, 
since  the  train  passes  along  the  latter. 

The  greatest  allowed  tensile  stresses  under  the  preceding  loads 
shall  be  : 

For  lower  chord  eye-bars  .     .  10,000  lbs.  per  sq.  in. 

"    main  brace  eye-bars  nearest  end  of  span      ....     10,000  "  "  "  " 

"     firsc  counter-brace 7-333  "  ''  "  " 

"    vertical  adjacent  to  end  of  span 7,333  "  "  "  " 

Other  tension  braces  to  be  proportioned  according  to  loca- 
tion between 7-333  and  10,000  "  "  "  " 

For  plate  hangers  on  floor-beams  (net  section)    ....       8,000  "  "  '•  " 

"    bottom  flanges  of  floor-beams  and  stringers  (net  section)  8,000  "  "  "  " 

■'    lateral  braces     .     .     .     .     , 12,000  "  "  "  " 

The  greatest  allowed  compressive  stresses  shall  be  : 
For  upper  chord  and  end  posts : 

Flat  ends.  Pin  ends. 

7,800                           ^             7,800  ,  . 

P= ^^72 .     •     .    /=  ^72 •     •     (i). 

50,000^^  3o,oooP^ 

For  intermediate  posts  at  centre  of  span,  a  reduction  of  20 
per  cent,  shall  be  made  from  the  preceding  values,  and  all 
other  intermediate  post  stresses  shall  be  proportioned  accord- 
ing to  location,  between  end  and  centre  values. 


39^  COMPLETE   DESIGN  OF  A    RAILWAY  BRIDGE. 

In  the  preceding  column  formulae,  "/"  is  pounds  per 
square  inch  ;  "/,"  length;  and  "r"  the  radius  of  gyration  of 
normal  section  in  direction  of  failure,  and  in  the  same  unit 
as  "  /." 

For  lateral  compression  braces  the  above  values  may  be 
increased  20  per  cent. 

For  top  flanges  of  stringers  and  floor-beams  the  greatest 
compressive  stress  shall  be  7,000  pounds  per  square  inch  of 
gross  section. 

The  greatest  mean  bearing  intensity  of  pressure  between  pins 
and  pin-holes  or  rivets  and  rivet-holes,  shall  be  12,000  pounds 
per  square  inch.  In  stringers  and  floor-beams  and  their  con- 
nections with  each  other  or  with  the  trusses,  this  value  shall 
be  reduced  25  per  cent. 

The  greatest  shearing  intensity  in  rivets  or  pins  shall  be 
7,500  pounds  per  square  inch,  and  in  stringers  and  floor- 
beams  and  their  connections  with  each  other  and  the  trusses, 
this  amount  shall  be  reduced  25  per  cent. 

The  greatest  bending  stress  in  the  extreme  fibres  of  wrought- 
iron  pins  shall  be  15,000  pounds  per  square  inch,  and  the  cen- 
tres of  pressure  shall  be  taken  at  the  centres  of  the  bearing 
surfaces. 

No  cast-iron  whatever  shall  be  permitted  in  any  part  of  the 
structure,  and  all  parts  shall  be  accessible  for  inspection  and 
painting. 

The  unsupported  width  of  any  plate  in  compression  shall 
not  exceed  thirty  times  its  thickness. 

The  pitch  of  rivets  in  compression  members  shall  not  ex- 
ceed sixteen  times  the  thickness  of  the  thinnest  plate  through 
which  the  rivets  pass. 

An  initial  stress  of  5,000  pounds  shall  be  added  to  that 
produced  by  the  vertical  loading  in  all  adjustable  tension 
members. 

These  meagre  specifications  are  sufificient  for  the  design 
when  it  is  premised  that  all  details  of  construction,  such  as 
eye-bar  heads,  connections,  etc.,  shall  be  consistent  with  the 
best  engineering  practice. 


COMPLETE  DESIGN   OF  A    RAILWAY  BRIDGE.  397 

Although  it  is  customary  to  add  the  total  stress  of  adjust- 
ment to  that  caused  by  the  vertical  loading,  in  adjustable 
tension  members,  the  practice  is  not  strictly  correct.  Just 
what  part  of  the  initial  stress  should  be  added  is  not  exactly 
determinate,  but  it  is  certainly  not  the  whole.  For  this  rea- 
son the  apparently  small  value  of  5,000  pounds  has  been  taken. 

As  the  widths  of  the  compression  members  depend,  to 
some  extent,  on  the  thickness  of  the  tension  members,  and 
as  the  design  of  the  latter  is  of  the  greatest  simplicity,  it 
conduces  to  the  greatest  convenience  to  begin  with  them. 
All  eye-bars  should  be  as  thin  as  considerations  of  resist- 
ance will  permit,  as  pin  bending  will  then  be  reduced  to  a 
minimum. 

By  taking  the  stresses  from  Fig.  i  of  PI.  II.,  and  subjecting 
them  to  the  preceding  specifications,  the  following  sections 
are  obtained  : 


Bars. 


Brace. 

Total  stress. 

A  llowed  stress. 

Sections. 

2 

45-553 

lbs. 

7,333 

lbs. 

per 

sq. 

in. 

2-4"  X     f" 

3 

167,800 

a 

10,000 

a 

u 

u 

" 

2—6    X  if 

5 

119,817 

a 

9,340 

Li 

u 

u 

(( 

2—6     X  l| 

7 

77,445 

(< 

8,670 

(< 

ii 

it 

<t 

2-5    X    1 

9 

47.654 

(< 

8,000 

a 

u 

u 

<( 

2-2"i    0 

10 

21,560 

(( 

7,333 

(i 

(< 

" 

a 

2-lf      " 

Lower  cho7-d. 

I  =  2 

131,520 

u 

10,000 

<( 

(< 

t( 

a 

2-5"xi"| 

3 

225,424 

a 

10,000 

(< 

li 

<< 

n 

4-5"xii 

4 

289,238 

<i 

10,000 

u 

li 

<( 

a 

6-5"  XI 

5 

322,220 

<< 

10,000 

(( 

u 

li 

<i 

6-5    xifV 

There  should  be  as  little  diversity  in  widths  of  bars  as  pos- 
sible, but  varying  thicknesses  within  standard  limits  are  easily 
produced  in  the  mill.  Again,  a  small  number  of  large  bars  is 
cheaper  to  produce  than  a  large  number  of  small  bars,  on 
account  of  the  smaller  number  of  pieces.  Hence  the  aim 
should  be  to  produce  large  pieces,  though  not  too  heavy  to 
be  handled  conveniently  in  the  shop. 

Before  designing  the  pins  the  intermediate  post  sections 


398 


COMPLETE   DESIGN   OF  A    RAILWAY  BRIDGE. 


Fig. 


sliould  be  determined.     These  posts  are  secured   to  pins  at 
each  end,  and  although  they  are  constrained,  by  some  extent 
in  the  vertical  plane  of  the  pin  axis,  it  is  only  slightly  so,  and 
they  should  be  considered  columns,  with  pin  ends  in  all  direc- 
tions.    They  will  each  be  built  of  two  channels  laced  in  the 
usual  manner.     The    depth  of  the  channel  is  an  important 
matter,  but  the  length  of  no  column  in  a  truss  should  exceed 
forty  times  its  least  diameter,  and  in    the    present  case  the 
depth  will  be  taken  at    ten   inches.       The 
channels  will   be    placed    as  shown    in   the 
I      figure  with  a  clear  separation  of  ten  inches. 
'^         Pin  plates  will   be  riveted  to  the  flanges 
I      of  the  channels  at  each  end,  through  which 
■       the    pin  will    pass,   leaving  the  axis  of  the 
latter  parallel  to  the  channel  webs  and  nor- 
mal to  the  planes  of  the  trusses.     The  least 
radius  of  the  post  section  will  be  parallel  to  the  pin   axis  and 
will  be  the  same  as  that  of  one  channel  about  an  axis  normal 
to    its  web,  or    about    3.9  inches.      The  length  of    the  post 
between  pin  centres  is  27  feet,  or  324  inches.     But   in  the 
plane  normal  to  the  truss  the  column    is  shortened  six  feet 
by  the  transverse   bracing  as  shown   in   Fig.    16  of    PI.   XII. 
Hence,  in  the  plane  of  the  pin  axis  /  -=-  r  —  252  -h-  3.9  =  65  ; 
and  in  the  plane  normal  to  the  preceding  r  =  5.8  .*.  /  h-  r  = 
324  -^  5.8  =  56.     As   the  post  is  considered  with  pin  ends   in 
all  directions  the  first  value  of  /  -f-  r  will  be  used. 

Eq.  (i)  then  gives  for  a  post  at  the  end/  —  6,840;  and  for 
one  at  the  centre  0.8  x  6,840  =  5,472.  Now  since  (6,840  — 
5,472)^  3  =  456; 

For  vertical  brace  4.  ./  =  6,840  —  456  =::  6,384  lbs.  per  sq.  in. 
"      6.  /  =  6,384- 456  =  5,928    "       "     "      " 
"     8.. z'' =  5,928  -  456  =  5,472    "       "     "      " 


The  initial  stresses  in  the  counters  intersecting  at  the  top 
of  vertical  brace  8  increase  the  stresses  in  that  member  8,000 
pounds.  The  preceding  quantities  then  give  the  following 
results : 


COMPLETE   DESIC7N  OF  A    RAILWAY  BRIDGE.  399 

Total  stress.  .       Allowed  stress.  Memhcr. 

Vertical  4 99,819  pounds. . .  .6,384  pounds.  ..  .2  —  10"  79  lb.  channels, 

6 66,193       "       5,928       "        2—  10    56" 

''        S 42,611        "       ....5,472       "       2  —  10'   48    "         " 

The  last  sectional  area  shows  a  material  excess  over  that 
required,  but  a  48-lb.  channel  is  about  the  lightest  rolled,  and 
this  excess  is  usually  found  in  the  centre  posts  of  trusses. 

The  lacing  on  these  posts  will  be  2\  x  -^^  placed  at  an 
angle  of  about  60°  with  the  post  axis. 

In  order  to  determine  the  lower  chord  pin  bending  some 

1 — 2  fir>soo 


\  1 

CO 

_  Centra ^^-EE^S""'' "^ 

45353 

Fig.   2. 

diameter  of  pin  must  be  assumed,  for  the  moment  at  the 
centre  of  the  pin  will  depend  partially  on  the  thickness  of  the 
pin  plates,  which  bear  against  the  pins.  A  diameter  of  4I 
inches  will  be  taken  ;  hence  each  inch  in  length  of  the  pin 
will  take  4,375  x  12,000  =  52,500  pounds.  It  will  be  seen 
hereafter  that  the  floor-beams  will  be  riveted  into  the  posts 
below  the  pins  in  such  a  manner  that  the  pin  plates  will  not 
only  carry  the  column  pressures  to  the  pin,  but  the  floor- 
beam  loads  also.  Hence,  in  determining  the  thicknesses  of 
bearing  areas  in   the  pins,  these  two  loads  and  the  pressures 

fiBsm  2 — 3 


due  to  initial  stresses  in  the  counter  rods  must  be  added. 
The  vertical  brace  8  is  the  only  post  subject  to  the  initial 
stresses  in  the  counters  and  the  vertical  component  of  each 
counter  at  its  top  is  4,000  pounds,  or  8,000  for  the  tuo. 


400  COMPLETE   DESIGN  OF  A    RAILWAY  BRIDGE. 

The  total  lower  chord  load  has  already  been  found  to  be  45,. 
553  pounds  in  the  case  of  brace  2.  This  maximum  lower  chord 
panel  load  will  not  usually  occur  with  the  greatest  post  stress, 
but  all  possible  cases  are  covered  by  combining  the  two.  The 
bearing  thickness  at  each  side  of  each  post  is  thus  found  to  be: 

For  brace  4 (99,819  +  45,553)  "^  52,500  x  2  =  1.4  inches. 

"       "      6 (66,193  +  45,553)  ^52,500    "     =1.06     " 

"      8 (34,611  +  45,553  +  8,000)^-52,500    "    =0.84     " 

By  regarding  the  principles  afTecting  pin  bending  as  de- 
veloped in  Art.  74,  it  will  be  found  that  the  arrangements  of 

o — 4  482W 

I  ■  I 

3  4S200 

-a \ ^J48aoQ 


53166C j.  =^ :^_ ^Co 

___Centre yS?__ 

Fig.  4. 

lower  chord  eye-bars  and  braces  shown  in  Figs.  2,  3,  4 
and  5  will  reduce  the  lower  chord  pin  bending  to  the  least 
amounts  possible. 

The  values  of  these  least  pin  moments  for  the  principal 

4S200 ,4 — 5 

'-—  ^. ,^ — ^  153700 

^) —  -1—  -  I  53700 

_    ^^P-i"-: liRS. 

Fig.  5. 

sections  are  found  to  be  as  follows,  and  they  can  be  verified 
by  remembering  the  thicknesses  of  the  eye-bars  (already  de- 
termined) and  the  fact  that  a  play  of  one-eighth  of  an  inch  is 
allowed  between  each  contiguous  pair  of  heads. 

Joint  1 — 2. 

The  section  of  plate  hanger  at  the  end  of  the  floor-beam  is 
shown  shaded,  and  the  distance  between  its  centre  and  that 


COMPLETE   DESIGN   OF  A   RAILWAY  BRIDGE.  40 1 

of  either  of  vertical  braces  2  is  3  inches.     Taking  moments 
about  the  centre  of  the  plate-hanger : 

Moment  about  Vert,  axis 65,800  x  1.5  =    98,700  in.  lbs. 

Hor.     "    22,800x3.0=    68,400"     " 

Herue  resultant  moment  ^   ^  (98,700)"  +  (68,400)^    =120,000"     "  .  (2). 

Joint  2 — 3. 

The  bearing  area  of  post  4  on  the  pin  is  shown  shaded.  In 
the  remaining  cases  it  will  be  assumed  that  the  adjacent 
lower  chord  panel  stresses  take  their  greatest  values  together, 
in  accordance  with  which  assumpt'^n  the  eye-bars  will  be 
stressed  for  this  joint  as  shown  in  Fig.  3.  The  force  78,400 
pounds  is  the  corresponding  stress  in  one  eye-bar  of  brace  3. 
The  tangent  of  the  inclination  of  the  latter  to  a  horizontal 
line  is  1.32,  hence  the  vertical  component  of  brace  3  (one  eye- 
bar)  is  (i  12,800  —  65,800)  1.32  =  62,040  pounds. 

The  moments  about  vertical  axes  are  : 

About  a 65,800  X  2.625  —    56,400  X  1.25  =  102,225  in.  lbs.    .    (3). 

"       b 65,800  X  4.0      —  112,800  X  2.0    =     37,600  "     " 

The  moment  about  a  horizontal  axis  through  c  is : 
62,040  X  1.5  =  93,060  in.  lbs. 

The  vertical  moment  about  c  is  the  same  as  that  about  b, 
hence  the  resultant  moment  about  c  is: 


|/ (37,600)" -4-  (93,060)- =  100,360  in.  lbs.       .      .  ....      (4). 

Joint  3—4. 

By  the  preceding  method  the  resultant  moment  of  the  in- 
clined brace  was  resolved  into  vertical  and  horizontal  com- 
ponents ;  in  this  and  the  following  cases,  however,  the 
resultant  moment  itself  will  be  taken.  As  before,  the  post 
bearing  is  shaded  in  Fig.  4.  The  head  of  the  counter  rod  c^ 
is  assumed  to  be  one  inch  thick.  The  secant  of  the  inclina- 
tion of  the  inclined  bar  to  a  horizontal  line  is  1.66.  Hence 
the  inclined  stress  is  (3  x  48.200—2  x  56,400)  1.66=  53,166 

pounds.     The  vertical  moments  are  then  as  follows : 
26 


402  COMPLETE  DESIGN    OF  A    RAILWAY  BRIDGE. 

About  a    .  .  .  48,200  X   1.2  =  57,840  in.  lbs. 

"        a  .  .  (f  12,800  —  96,400)  X   2.94  =  48,216  "■      " 

"        b  .  .  .  .  48,200  X  3.5  —  16,400  X  6.44  =  63,100  "      " 

The  inclined  moment  about  b  is ; 

53,166  X  2.31  =  122,800  in.  lbs. 

In  Fig.  6  ab  is  normal  to  the  inclined  brace,  and  represents 
122,800  inch-pounds  by  scale,  while  be  is  vertical,  and  repre- 
sents 63,100  inch-pounds. 

Hence  the  resultant  moment  about  b  is  represented  by  ac, 
and  has  the  value  : 

R  =  97,500  inch  pounds (5). 

S^otn^  4 — 5. 

The  thickness  of  the  head  of  the  centre  rod  c  is  1.5  inches; 
and  the  inclined  eye-bar  stress  of  brace  7  is 

(3  X   53,700  —  3   X  48,200)  1.66  =  27,500  pounds. 

The  vertical  moments  are  as  follows : 

About  (z'  48,200  X  1,155  =   55,671  in.  lbs.  (6). 

"  a    2(53,700  —  48,200)  X  2.875  —   31,625"  " 

"     i    2  (53.700  —  48,200)  X  4.03  —  48,200  X  1. 155  =  —  11,340  "  " 

"  ^  3  (53,700  -  48.200)  X  3.7  -11,340       =   49»7oo  "  " 

The  inclined  moment  about  d  is 

27,500  X  2.6  =  71,500  in.  lbs. 
Hence  the  resultant  moment  about  d  is,  by  Fig.  7: 

R  =  57,000  inch-pounds '  (7). 

In  addition  to  the  preceding,  the  moments  of  the  greatest 
vertical  components  of  the  inclined  eye-bar  stresses  about  the 
centres  of  the  post  bearings  should  be  examined.  It  is  here 
unnecessary  to  go  into  all  these  in  detail.  The  greatest  oc- 
curs at  joint  3 — 4.    The  vertical  component  of  the  maximum 


COMPLETE    DESIGN   OF  A    RAILWAY  BRIDGE. 


403 


•Stress  is  ( 1 19,817  ~  2)  0.8  =  47,930  lbs. 
in  question  is  : 


Hence  the  moment 


47,930  X  2.31  —  1 10,700  in.  lbs. 


(8). 


The  preceding  results  show  that  while  it  is  quite  unneces- 
sary to  take  moments  at  the  centre  of  all  bearings,  a  thor- 
ough examination  of  the  lower  chord  joints  must  be  made  in 
order  to  find  the  greatest  moments. 

Eq.  (2)  gives  the  greatest  resultant  moment  of  120,000 
inch-pounds.  Hence  a  wrought-iron  pin  4,375  inches  in  diam- 
eter will  be  sufificient  to  meet  the  requirements  of  the  speci- 
fications. But  since  the  bars  in  braces  3  and  5  are  6  inches 
wide,  and  since  the  eye-bar  heads  are  no  thicker  than  the 
bodies  of  bars,  the  requirement  of  12,000  pounds  per  square 
inch  bearing  pressure  against  pins  cannot  be  met  by  a  less 
diameter  than  5  inches  in  the  case  of  brace  3.  For  a  reason 
that  will  appear  hereafter,  the  diameter  of  the  large  pins  will 
be  taken  at  5f  inches. 


Fig.  6. 


Fig.  7. 


Fig.  8. 


In  the  cases  of  braces  7  and  9  a  much  smaller  pin  may  be 
used,  and  while  it  is  not  economy  in  the  shop  to  have  a  large 
number  of  pin  diameters,  two,  or  even  three,  are  not  too  many 
for  a  span  of  this  length.  The  cosine  of  the  inclination  of 
brace  7  to  a  horizontal  line  is  0.61,  hence  the  horizontal 
component  of  the  greatest  stress  in  one  of  its  eye-bars  is 
(77,445  -^  2)  X  0.61  —  23,620  pounds.  It  will  be  seen  here- 
after that  the  side  plates  of  the  upper  chord  panels  3  and  4 
will  be  0.5  inch   thick,  and  since  4^\   x   12,000  -^  2  =  25,000, 


404 


COMPLETE    DESIGN   OF  A    RAILWAY  BRIDGE. 


it  appears  that  with  a  pin  diameter  of  a^-?^  inches  no  thicken- 
ing plates  at  pin-holes  D,  E,  F,  and  M  will  be  needed.  It 
will  be  found  that  the  thickness  of  bearing  plates  at  the  top 
of  brace  6  must  be  \\  inch,  and  the  clearance  at  each  side  of 
eye-bar  head  (between  i-inch  side  plate  of  chord  and  pin  plate 
of  post)  will  be  about  y^  inch.  The  vertical  component  of 
eye-bar  stress  (for  brace  7)  will  be  (77,445  -h  2)  x  0.8  =  30,980 
pounds.     Hence : 

Pin  moment  at  centre  of  eye-bar  head  =  23,620  x  0.875  =  20,670  in.  lbs. 
Vertical  "  "  post  pin  plate  =  23,620  x  1,875=44,290"     " 

Inclined  "  "  "  =  38,720  x  i.oo    =  38,720  "     " 

The  resultant  of  the  last  two  moments  is  shown  by  Fig.  8 
to  be : 

R  =  37,000  inch-pounds (9). 

The  moment  of  the  greatest  vertical  component  in  brace  7 
at  the  bottom  of  post  8  is : 

30,980  X   2,6  =  80,550  inch-pounds.      .     (10). 

Eqs.  (6),  (7),  (9),  and  (10)   show  moments   far   below  the 

allowed    resisting  capacity  of   a  4^  wrought-iron   pin,   i.  e., 

108,000  inch  pounds. 

Hence,  at  C,  L,  K,  and  J  5|  inch  pins  will  be  used,  while  at 

D,  E,  and  I,  ^^^  pins  will  be  taken. 

Before  determining  the  diameters  of  the  pins  in  the  inclined 

end  post,  it  will  be  necessary  to  fix 
the  sections  of  that  member,  and  it 
will  be  convenient  to  find  those  of 
the  upper  chord  at  the  same  time. 
As  each  upper  chord  panel  is  a 
beam  of  considerable  span,  carry- 
ing its  own  weight,  the  depth  should 
not  be  small,  and  it  will  be  taken 
at  eighteen  (18)  inches.  The  radius 
of  gyration  of  the  normal  section 

about  a  horizontal  axis  through  its  centre  of   gravity  must 

first  be  found. 


% 


— ^ G — -^^--H 


ri- 


-.t- 


jL 


Fig.  9. 


COMPLETE   DESIGN   OF  A    RAILWAY  BRIDGE.  405 

As  the  upper  chord  stress  in  panels  3  and  4  is  about 
322,000  pounds,  the  area  of  that  panel  section  will  not  be  far 
from  44  square  inches.  Fig.  9  represents  a  trial  section  of 
that  area. 

AB  is  a  21    X  ^»g  inch  cover  plate. 

C  and  D  are    3x3      "     21  lb.  angles. 

£       a      P      u  5x3  "50 

6^    "    //  "     18  X    1       "     side  plates. 

The  5-inch  legs  of  E  and  F  are  horizontal.     The  centres 

of  gravity  of  C  and  D  are  0.9  inch  from  lower  surface  of  AB, 

and  those  of  E  and  F  are  0.9  inch  from  lower  surface  of  the 

horizontal   5-inch  legs.     Static   moments  about  a  horizontal 

line  through  the  centre  of  gravity  of  the  section  of  AB  give: 

3x3  angles     .     .     .     2  x  2. i   x     1.2  =       5.04 

5x3      "  ...     2  X   5.0  X   17.4  =  17400 

Side  plates        ...  18  x     9.3  =  167.4 

Total 346.44 

Hence,  346.44  -4-  44  =  7.9  inches ;  or  the  centre  of  gravity 
g  of  the  e?itire  section  is  y.6  inches  from  the  lower  surface  of 
AB.  In  such  computations  some  dimensions  are  taken  a 
little  full  because  adjacent  surfaces  do  not  have  mathemati- 
cal contact. 

The  elements  of  the  moment  of  inertia  of  the  section 
about  the  horizontal  axis  GH  through  g  take  the  values  : 

Cover  plate 11. 81   x  7.9-  =  737-o6 

,  (    2        X  2.0   =      4.0 

3  X   a  angles        \  ,^.,  „„ 

•^        -^       ^  (    4.2     X  6.7-  =  188.54 

r     y^    .         u  j      2  X    4^    =         9.0  • 

^       ^  (  lO.O     X  9.52  =  902.5 

Side  plates     ....      J  18  x   Is^  -  1^=486.0 

(  18    X    1.42  =     35.28 

Moment  of  inertia     ....     2,362.38 

The  moment  of  inertia  of  AB  about  a  horizontal  axis 
through  its  own  centre  of  gravity  is  so  small  that  it  has  been 
neglected.     The  3x3  and  5x3  angles  each  have  a  moment 


40^  COMPLETE  DESIGN   OF  A    RAILWAY  BRIDGE. 

of  inertia  of  2  about  a  horizontal  axis  through  the  centre 
of  gravity  of  each  respective  section.  The  least  radius  of 
gyration  about  a  horizontal  axis  for  the  entire  section  then 
becomes : 

V 2,362.38  -^-  44  =  7-33  inches. 

The  panel  length  is  20.55  ^^-     Hence 

l^r  =  20.55  X  12  -f-  7.33  =  33.7. 

The  preceding  value  in  Eqs.  (i)  gives: 

Flat  ends.  Pin  ends, 

p  =  7,620  lbs.  per  sq.  in.  p  —  7,5 10  lbs.  per  sq.  in. 

For  one  pin  and  one  fiat  end /  =  (7,620  +  7,510)  -^-  2 

=  7,565  lbs.  per  sq.  in. 

The  upper  chord  will  be  continuous  after  the  bridge  is 
erected,  but  the  extremities  will  be  hinged  at  the  upper  ends 
of  the  inclined  end  posts  in  the  manner  shown  in  Fig.  4  of 
PI.  XI.  Hence  upper  chord  panel  i  will  have  one  pin  end 
and  one  flat  end  ;  all  other  panels  will  be  flat  end  columns. 
The  upper  chord  sections  will  now  be  as  follows : 

Upper  chord  i. 
Required  area  =  225,424  -=-  7,565  =  30.0  sq.  ins. 

1  —  21  X  y'L  inch  cover  plate.  9.2  "  " 
2—3x3  "  18  lb.  angles  3.6  "  " 
2—5x3      "     30  "       "         6.0  "      " 

2  —  18  x    I      "    side  plates.  13.5  "      " 


Total 32.3  "  " 

Upper  chord  2. 

Required  area  =  289,238  -i-  7,620  =  38.0  sq.  ins. 

1  —  21   X  ^'g-  inch  cover  plate  .  9.2  "  " 

2  —     3x3     "      18  lb.  angles.  3.6  "  " 
2  —      5    X     3      "       36  "          "            7.2    "  " 

2  —  18  X    i     "     side  plates.  .18.0  "  " 

Total 38.0  "  " 


COMPLETE   DESIGN  OF  A   RAILWAY  BRIDGE.  407 

Upper  chord  3  and  4. 
Required  area  =  322,220  ^  7,620  =      42.3  sq.  ins. 

1  —  21  X  y"?^  inch  cover  plate  =  9.2    " 

2  —     3x3     "      18  lb.  angles     3.6    " 

2  -     5  X    3     "      47  "        "  9-4    " 

2  —  18  X  y9^    "      side  plates.  20.25     " 


Total 42.45     " 

The  end  post  bears  on  pins  at  top  and  bottom  ;  hence  it  is 
a  pin-ended  column.  It  is  about  408  inches  long,  and  it  will 
be  most  convenient  to  take  its  depth  identical  with  that  of 
the  upper  chord,  or  18  inches.  The  radius  of  gyration  may 
then  be  taken,  as  before,  at  7.33  inches.  Hence,  i  ^  r  =  408 
-r-  7.33  =  55.7.     The  second  formula  of  Eq.  (i)  then  gives: 

/  =  7,070  lbs.  per  sq.  in. 

Inclined  end  post. 

Required  area  =  217,750  -^  7,070  =  30.8  sq.  ins. 

1  —  21   X  yig-  cover  plate  ....  9.2  "      " 

2  —  3  X  3  18  lb.  angles  .  .  3.6  "  " 
2  —  5  X  3  30  "  "  .  .  6.0  "  " 
2  —  18  X    I  side  plates. ...  13.5  "      " 


Total 32.3  "      " 

All  these  actual  areas  agree  sufificiently  near  in  character 
and  amount  with  the  trial  section  to  make  re-computations  of 
the  radius  of  gyration  quite  unnecessary.  A  very  little  ex- 
perience makes  such  a  result  possible  in  all  ordinary  cases. 
A  very  close  but  approximate  rule  for  all  box  or  semi-closed 
sections  like  those  just  considered,  is  to  take  the  radius  of 
gyration  at  four-tenths  (0.4)  the  depth  of  the  side  plates.  In 
the  present  case  it  would  make  r  =  0.4  x  18  =  7.2  inches,  while 
the  exact  value  is  7.33  inches. 

In  building  a  section  such  as  these,  the  angles  C  and  D, 
Fig.  9,  should  be  made  as  light  as  possible,  in  order  that  the 
cover  plate  AB  may  be,  to  a  considerable  extent,  balanced  by 
the  heavy  angles  £,  F.     In  this  manner  the  centre  of  gravity, 


408  COMPLETE  DESIGN  OF  A    RAILWAY   BRIDGE. 

g,  of  the  section  may  be  brought  down  sufficiently  near  to 
the  mid  depth  to  give  all  the  space  needed  inside  the  chord 
for  the  eye-bar  heads,  if  the  pin  axis  should  be  made  to  pass 
through  g,  at  the  same  time  there  is  gained  the  incidental  but 
important  advantage  of  an  increased  moment  of  inertia.  If 
the  chord  were  subject  to  no  bending  from  its  own  weight, 
the  axis  of  every  pin  should  pass  through  the  centre  of  gravity 
of  the  section.  It  has  been  shown  in  Art.  69  that  this  flexure 
cannot  be  satisfactorily  neutralized  by  the  direct  stress,  par- 
ticularly if  the  chord  is  continuous,  as  in  the  present  case.  It 
is  best,  therefore,  to  reduce  the  bending  stresses  by  making 
the  chord  depth  as  great  as  possible.  For  these  reasons  it 
was  taken  at  eighteen  (18)  inches.  If  the  panels  were  non- 
continuous  the  greatest  stress  per  sq.  in.  in  the  exterior  fibres 
of  panels  3  and  4  would  be  : 

j^      Md       88,800  X  10.4 

K  =  ~j—  =  — '-^ p =  300  lbs. 

/  2,362  ^^ 

Now,  when  it  is  remembered  that  the  chord  is  continuous 
it  is  evident  that  flexure  may  be  neglected  in  the  sections 
found.  This  point,  however,  should  always  receive  careful 
attention. 

In  the  present  case,  //le  axes  of  pins  in  the  upper  chord  and 
end  posts  zvill  be  placed  eight  (8)  inches  from  the  line  AB,  thus 
allowing  a  small  counter  moment  from  the  direct  stress  due 
to  a  lever  arm  of  0.4  inch. 

A  compression  member  with  the  same  degree  of  end  con- 
straint in  all  directions  ought  to  have  equal  capacity  for  re- 
sistance in  all  directions.  If  the  radius  of  gyration  be  taken 
in  different  directions  about  g.  Fig.  9,  for  the  different  sec- 
tions as  formed,  it  will  be  found  that  this  condition  is  fulfilled. 

Finally,  the  unsupported  width  of  any  plate  in  compression, 
measured  transversely  between  rivet  heads,  should  not  be 
more  than  about  thirty  times  the  thickness.  An  examination 
of  the  sections  will  show  that  this  condition  also  has  been 
fuffilled. 

The  details  about  the  pin  bearings  at  the  upper  and  lower 
ends  of  the  inclined  end  post  may  now  be  considered. 


COMPLETE   DESIGN  OF  A    RAILWAY  BRIDGE.         4O9 

Fig.  4  of  PL  XI.,  shows  the  detail  at  the  upper  end  of  the 
inclined  end  post.  The  end  panel  of  the  upper  chord  does 
not  rest  its  extremity  immediately  against  the  upper  end  of 
brace  i,  but  they  are  separated  along  the  line  ab  by  about 
the  distance  of  |  inch,  and  each  bears  directly  against  the 
end  pin.  The  diameter  of  the  latter  is  taken  by  trial  at  5^ 
inches.  By  the  specifications  the  bearing  value  of  this  pin 
against  a  one-inch  plate  is  5|  x  12,000  =  61,500  pounds. 
Hence  for  plates  J,  y^^,  |,  \  and  ^  inch  the  bearing  values  will 
be  as  follows : 

61,500  X   ^  =  15,375  pounds. 
"        X  tS  =  19,220       " 
"        X  I  =  23,060       " 
X   J    =  30,750       " 

"  X^=:   34,596  " 

The  arrangement  of  thickening  plates  for  upper  chord  i  is 
clearly  shown  by  Fig.  4 ;  there  is  a  half-inch  plate  inside  and 
next  the  fths  web  and  a  |  jaw  plate  inside  the  half-inch  thick- 
ener. Against  the  web  outside  is  a  ^^g.  inch  thickener.  The 
total  bearing  thickness  is  then  |  +  i  4  |  +  ^^  =  if|  inches. 
Hence  61,500  x  \\\  =  1 1 1,500  pounds.  The  half  of  the  stress 
in  upper  chord  i  is  112,712  pounds  and  the  two  quantities 
are  sufificiently  near  in  amount.  All  rivets  about  the  joint 
are  f  inch  in  diameter.  The  shearing  resistance  of  one  rivet 
at  7,500  pounds  per  sq.  in.  is  3,300  pounds,  while  the  bearing 
values  against  the  various  plates  are : 

I  rivet  against  -^-^  plate  =  2,800  pounds. 
"     "  "         I       "      =  3,400        " 

"     "  "        I      "      =  4,500        " 

The  total  bearing  pressure  against  the  jaw  and  thickening 
plates  is  23,060  +  30,750  +  34,596  =  88,406  pounds,  and  there 
are  21  rivets  through  those  plates,  as  shown  in  Fig.  4.  Ap- 
plying the  bearing  and  shearing  values  given  above  to  the 
number  and  distribution  of  rivets  located  in  that  figure,  it 
will  be  seen  that  there  is  a  little  excess  of  both  those 
resistances. 


4IO  COMPLETE  DESIGN  OF  A    RAILWAY  BRIDGE. 

The  same  figure  shows  the  number  and  distribution  of  both 
rivets  and  thickening  plates  at  the  upper  end  of  the  incHned 
end  post.  There  is  a  jj  inch  jaw  plate  outside,  then  a  half- 
inch  thickener  and  a  f^  plate  between  that  and  the  web. 
There  is  also  a  quarter-inch  thickener  inside.  The  amount  of 
bearing  thickness  is  thus  the  same  as  for  the  upper  chord. 
An  examination  of  the  number  and  location  of  the  rivets  will 
show  that  there  is  again  a  little  excess  in  the  bearing  and 
shearing  resistances.  The  pitch  of  rivets  in  the  immediate 
vicinity  of  the  joint  is  three  (3)  inches,  in  all  other  parts  of  the 
upper  chord  and  end  posts  it  will  be  six  (6)  inches. 

The  preceding  arrangement  is  for  one  side  of  the  chord  and 
end  post,  since  both  sides,  of  course,  are  alike.  As  Fig.  4 
shows,  the  pin  passes  through  the  jaw  plates  only.  The  four 
(4)  jaw  plates  hold  the  post  and  upper  chord  securely  to- 
gether in  case  of  any  derailment  or  other  accident  tending  to 
knock  the  end  post  out  of  place.  They  are  further  reinforced 
by  the  light  f  inch  cover  plate  shown  at  a.  The  latter  also 
performs  an  important  of^ce  in  transferring  upper  lateral 
loads  to  the  end  posts.  One  portion  of  it,  or  both,  must,  of 
course,  be  riveted  in  the  field.  It  will  be  observed  that  each 
jaw  plate  has  a  "play"  or  clearance  of  \  inch,  to  provide  for 
imperfections  of  workmanship  and  secure  ready  erection. 
The  figure  shows  what  rivets  must  be  countersunk,  both  out- 
side and  inside. 

The  eye-bars  of  brace  3  lie  adjacent  to  the  interiors  of  the 
upper  chord  and  end  post,  while  those  of  brace  2  are  inside 
of  the  first.  Assuming  that  the  greatest  stresses  in  those 
braces  occur  together  (which  is  a  small  error  on  the  side  of 
safety),  the  end  pin  will  be  subjected  to  the  bending  moments 
shown  in  Fig.  10.     The  component  moments  are  as  follows: 

„      ,                          167,800  .      ,, 

For  brace  3-     • x  1.71  =  143.470  i"-  lbs. 


"        "      2-     •     •    "^^'^^^  X  2.90  =    66,053  "      " 
2 


217,750  X  0.62  —    67,500  "      " 


COMPLETE   DESIGN  OF  A    RAILWAY  BRIDGE.         4II 

The  latter  moment  arises  from  the  fact  that  the  upper 
chord  and  end-post  bearings  have  their  centres  separated  by 
0.62  inch. 

In  the  figure,  be  is  normal  to  brace  3,  and  ac  is  horizontal^ 
while  ad  is  normal  to 
brace  i.  Hence  db  is  the 
resultant  moment  of  220,- 
000  inch  pounds.  A  pin 
51  inch  in  diameter  will  a 
little  more  than  supply 
the  required  resistance 
with  K—  15,000,  as  Eq. 
(i)  of  Art.  74  demon- 
strates,  or  as  may  more 

simply  be  found  by  reference  to  any  reliable  table  of  pin  mo- 
ments. The  thickening  plates  and  rivets  just  found  will  now 
be  a  very  little  excessive,  but  they  will  be  retained. 

These  large  rounds  frequently  vary  in  standard  sizes  by 
quarter-inches,  and  a  5J  inch  diameter  may  be  turned  to  5| 
with  little  waste. 

Fig.  7  of  PI.  XI.  shows  the  lower  end  of  the  inclined  end 
post  with  the  number  and  location  of  the  |  rivets  and  thick- 
ening plates.  The  operation  of  designing  them  is  precisely 
similar  to  those  already  employed,  and  they  will  not  now  be 
repeated.  An  examination  of  the  plates  and  rivets  in  con- 
nection with  the  preceding  values,  will  show  that  the  shear- 
ing and  bearing  resistances  of  both  the  rivets,  plates,  and  5=^ 
inch  (assumed)  pin  required  by  the  specifications  are  secured. 
The  line  ab  is  2\  inches  below  the  centre  of  the  pin-hole, 
giving  about  2  inches  of  solid  metal  below  the  pin. 

Figs.  8,  9  and  10,  of  PI.  XI.,  show  two  elevations  and  a 
plan  of  the  pedestal  at  the  lower  extremity  of  each  end  post. 
The  centre  of  the  pin-hole  is  taken  six  (6)  inches  above  the 
bottom  I  inch  plate.  The  figures  show  with  perfect  clearness 
the  arrangement  of  the  various  parts.  The  4  and  19  inch 
spaces  give  ample  clearance  for  the  sides  of  the  end  post 
which  enter  them. 

The  vertical  component  of  the  end-post  stress  is  172,820 


4^2  COMPLETE   DESIGN  OF  A    RAfLWAY  BRIDGE. 

pounds.  The  total  bearing  thickness  under  each  half  of  the 
pin  is  I  +  ^  +  8  =  if  inches.  But  64,500  x  !«  =  104,610;  or 
greater  than  172,820  -^  2  ■=  86,410.  Hence,  ample  bearing 
surface  at  12,000  pounds  per  square  inch  is  secured.  Now 
since  the  half  of  the  end-post  bearing  is  at  the  centre  of  each 
of  the  4  inch  spaces,  it  may  at  first  sight  appear  as  if  either 
equal  bearing  areas  ought  to  be  found  each  side  of  those 
spaces,  or  as  if  all  ought  to  be  on  one  side.  But  it  is  better 
to  mass  the  metal  as  much  as  possible ;  at  the  same  time  the 
weight  should  be  distributed  somewhat  on  the  f  inch  plate. 
The  arrangement  shown  accomplishes  these  results  and  gives 
a  little  excess  of  bearing  area.  The  5  by  3  inch  angles  are 
but  15  inches  long,  while  the  |  inch  bottom  plate  is  24  by  36 
inches. 

The  bearing  thickness  at  a  is  1.25  inches;  hence  the  up- 
ward pressure  at  that  surface  is  64,500  x  1.25  =  80,625 
pounds. 

The  4  inch  space  gives  about  \\  inch  total  clearance  for  the 
side  of  the  end-post  and  the  eye-bar  (i|  inches  thick)  of  lower 
chord  panel  i,  or  ^  inch  each  for  the  three  clearance  spaces 
thus  formed. 

The  pin  moment  about  the  centre  of  the  end-post  bearing 
is: 

80,625  X  1.875  inches  =  151,171  m/; /<^j\     .     .     .     (11). 

Again,  taking  moments  about  the  centre  of  the  angle  bear- 
ing b,  there  are  two  moments  with  horizontal  axes  but  with 
opposite  signs  formed  by  the  upward  pressure  at  a,  and  the 
half  vertical  component  in  the  inclined  end  post,  thus: 

-f-  80,625  X  4.81  =  -^  387,806  in.  lbs. 
—  86,413  X  3.00=  —  259,230  "     " 


Resultant^  -I-  128,576  "     " 

The  stress  in  the  5  x  i|  inch  eye-bar  of  lower  chord  i  has 
the  following  moment  about  a  vertical  axis  passing  through 
the  centre  of  b  : 

65.760  x  1.14  =  74,970  in.  lbs. 


COMPLETE   DESIGN  OF  A    RAILWAY  BRIDGE.         413 
Hence  the  resultant  moment  about  the  centre  of  <^  is : 


-v/(i28,576)^ +  (74,970)- =  150.000  ?/'/.//''i-.     .     .     .    (12). 

As  the  moment  (11)  is  greater  than  (12)  and  far  less  than 
the  resisting  capacity  of  the  assumed  5  J  inch  pin,  the  latter 
diameter  will  be  retained.  A  smaller  pin  would  give  sufficient 
bending  resistance,  but  would  necessitate  additional  metal  in 
the  thickening  plates,  and  would  increase  the  variety  in  pins 
and  pin-holes. 

It  is  frequently  desirable  to  hang  one  pair  of  eye-bars 
(either  braces  2  or  3)  outside  of  chord  and  end  post  at  the 
upper  end  of  the  latter.  In  such  a  case  the  angle  flanges  at 
i,  Fig.  4,  PI.  XL,  would  be  cut  away,  and  more  rivets  would 
need  to  be  countersunk  about  the  pin-hole  on  the  outside  of 
the  outer  jaw  plate. 

In  the  present  instance,  however,  the  pin  necessary  at  the 
upper  chord  end  is  but  little  different  from  those  required  by 
braces  3  and  5.  Hence,  for  the  sake  of  uniformity,  the  pins 
at  B,  C,  A,  L,  K  and/,  will  be  given  a  diameter  of  5|  inches, 
while  the  others  are  4Y^g^  inches. 

The  pin-plates  at  the  upper  and  lower  end  of  the  intermediate 
posts  will  now  be  found. 

It  will  be  assumed  that  the  maximum  post  stress  and  the 
greatest  panel  load  occur  together.  This  is  not  possible,  but 
it  is  difficult  to  determine  the  exact  maximum  load  on  the 
lower  pin-plate,  and  the  assumption  involves  a  safe  error.  It 
will  farther  be  assumed  that  the  greatest  panel  load  for  the 
intermediate  posts  is  the  same  as  the  greatest  load  on  brace 
2.     This  also  involves  a  slight  safe  error. 

In  consequence  of  these  assumptions  and  the  additional 
fact  that  the  smaller  part  of  the  load  in  each  lower  pin-plate 
is  the  panel  moving  load,  no  addition  for  impact  will  be  made 
in  fixing  the  thickness  of  the  pin-plates. 

The  manner  of  supporting  the  ends  of  the  floor-beams  is 
clearly  shown  in  Figs.  14,  15  and  16,  PI.  XI.  They  are  built 
into  the  posts  below  the  pin.  The  channels  are  continued 
30  inches  below  the  centres  of  the  pin-holes,  and  a  4  by  4  inch 


414  COMPLETE  DESIGN  OF  A    RAILWAY  BRIDGE. 

36  pound  angle  is  riveted  to  each  as  shown  2X  a  a  Fig.  14. 
The  end  stiffeners  of  the  floor-beam  (Fig,  2,  PI.  XI.)  are 
brought  against  these  latter  and  riveted  fast  to  them  in  erec- 
tion. The  number  of  rivets  required  for  this  connection  will  be 
found  later  on.  In  order  to  freely  admit  the  end  stiffeners 
of  the  floor-beam,  the  10  inch  channels  of  the  post  will  be 
separated  10  inches. 

Vertical  Brace  4. 

The  top  pin  plates  will  carry  99,819  lbs. 
"     bottom  "     "      "       "      99,816  +  45,553  =  I45>372  lbs. 

Since  5|  x  12,000  =  64,500  pounds,  the  total  thickness  of 
bottom  pin  plates  will  be  145,372  -^-  64,500  =  2.25  inches  ;  and 
since  the  shearing  resistance  of  one  f  inch  rivet  is  3,300 
pounds,  the  total  number  of  rivets  in  the  channel  flanges  will 
be  145,372  -=-  3,300  =  44.  The  lower  part  of  Fig.  14,  PL  XL 
shows  the  required  arrangement  of  pin  plates  and  rivets 
There  are  three  |  inch  outside  pin  plates  on  each  side  of  the 
post.  The  rivets  about  the  pin-hole  on  the  outside  will  be 
countersunk  in  order  that  the  eye-bar  of  brace  3  may  lie 
close  against  the  post. 

The  total  thickness  of  pin  plates  at  the  top  of  the  post  will 
be  99,819  -=-  64,500  =  1 1^6  inches,  and  the  total  number  of 
rivets,  99,819  -^  3,300  =  30.  The  upper  part  of  Fig.  14,  PI, 
XI.  shows  the  required  arrangement  of  pin  plates  and  rivets. 
As  the  total  number  of  the  latter  must  be  divided  by  4,  32 
rivets  are  used.  There  is  one  |  inch  outside  pin  plate  and 
one  yV  inch  inside  plate  riveted  to  the  former.  The  object  of 
placing  the  latter  inside  is  to  keep  the  upper  chord  as  narrow 
as  possible. 

Vertical  Brace  6. 

The  top  pin  plates  will  carry  66,193  lbs. 
"     bottom"  "      "       "       66,193  +  45,553  =  1 11,746  lbs. 

The  total  thickness  of  bottom  pin  plate  will  be  111,746^ 
64,500=  if   inches,  and   that   of    the   upper  66,193  —  50,250 


COMPLETE   DESIGN  OF  A    RAILWAY  BRIDGE.         415 

=  i/g  inches,  since  the  upper  pin  is  4y^e  inches  in  diameter 
and  4y\  x  12,000=  50,250  pounds.  The  total  numbers  of  \ 
rivets  below  and  above,  respectively,  are  1 1 1,746  -r  3,300  =  34 
and  66,193  -^  3,300=  20.  Fig.  15,  PI.  XL,  shows  the  required 
pin  plates  and  rivets.  At  the  bottom  there  is  a  \  inch  plate 
next  to  the  channels  and  a  half-inch  plate  outside.  At  the 
top  there  is  a  |  inch  plate  outside  and  a  ^g  i^ich  plate  inside, 
as  shown. 

Vertical  Brace  8. 

Four  adjustable  ties  meet  the  upper  extremity  of  this  post, 
and  it  has  already  been  shown  that  each  tie  adds  4,000 
pounds  to  the  post  stress.      Hence,  the 

top  pin  plates  will  carry  34,611   x  16,000  =50,611  lbs. 

bottom         "      "       "        34,611  -f  16,000  +  45,553=96,164    " 

The  total  thickness  of  bottom  pin  plates  will  be  96,164  -=- 
50,250=  \\  inches;  and  that  of  the  top  plates  50,611-^ 
50,250  =  I  inch.  The  total  numbers  of  rivets  required  are 
96,164^  3,300  =  29  and  50,611  ^3,300=  16.  Fig.  16,  PL 
XI.,  clearly  shows  the  arrangement  of  plates  and  rivets. 
There  are  more  rivets  show^n  at  the  top  than  is  necessary  tor 
bearing  or  shearing  alone,  for  the  reason  that  the  notch  a 
must  be  cut  out  of  one  channel  to  let  the  counterbraces  9 
take  hold  of  the  pin  inside  the  post,  as  there  is  not  room 
enough  outside. 

Upper  Chord  Joints  and  Thickening  Plates. 

It  will  readily  be  seen  that  the  pin-hole  at  the  joint  point 
between  upper  chord  i  and  2  is  the  only  one  needing  a  thicken- 
ing plate.  The  greatest  tension  in  an  eye-bar  of  brace  5  is 
119,817  -^  2  =  59,909  pounds,  and  the  sine  of  its  inclination 
to  a  vertical  line  is  0.61.  Hence,  its  horizontal  component  is 
59,909  X  0.61  =  36,544  pounds.  The  thickness  of  the  side 
plates  of  upper  2  is  \  inch;  hence.  \  x  I2,000  x  5|  =  32,250 
pounds.  A  little  over  4,000  pounds,  then,  is  all  that  need 
be   resisted   by  a    thickening    plate.     This    might    safely  be 


4i6 


COMPLETE   DESIGiX   OF  A    RAILWAY  BRIDGE. 


neglected,  but  the  y^g  joint  plate  shown  by  Fig.  13,  PI.  XL, 
will  be  extended  to  cover  the  pin-hole. 

Precisely  the  same  operation  shows  that  no  thickening 
plates  are  needed  at  the  other  pin-holes. 

There  will  be  joints  in  the  upper  chord  as  near  as  possible 
to,  and  on  the  left  of  the  pin-holes  at  6",  D  and  E  of  Fig.  i, 
PI.  II.,  and  at  corresponding  points  in  the  other  half  of  the 
truss. 

These  joints  are  formed  as  shown  at  Fig.  13,  PI.  XI.  At 
C  the  joint  will  be  12  inches  from  the  centre  of  the  pin  hole. 
It  is  formed  by  riveting  top  and  bottom  and  side  plates  to 
the  chords,  as  shown.  All  the  joints  are  formed  precisely 
like  this,  except  that  in  the  other  cases  the  y^^  plate  between 
the  angles  extends  each  side  of  the  outer  one,  as  shown  on 
the  left  only. 

Latticing  and  Batten  Plates. 

The  dimensions  of  latticing  and  batten  plates  are  matters 
of  judgment  and  experience.  Evidently  no  segment  of  a 
column  between  lattice  points  ought  to  be  less  in  resistance 
per  square  unit  of  section  than  the  column  as  a  whole,  but 
experiments  are  yet  lacking  to  give  quantitative  results. 
Single  latticing  with  centre  lines  making  angles  of  60°  with 
the  axis  of  the  member  will  be  used  here. 

On  the  under  side  of  the  upper  chord  and  end  post  the 
lattice  bars  will  be  4  inches  by  |,  and  each  end  will  be  held  by 

two  rivets.     Fisr.  1 1  shows 


/oo/ 

\oo\ 

v 

/oo/ 

\p 

-/ / 

/    / 

^ 

\^ 

// 

IT 

/op/ 

VoN 

7o=r 

_ 

Fig.  II. 


this  latticing.  It  will  weigh 
about  eight  (8)  pounds  per 
lineal  foot  of  member.  The 
two  battens  (one  at  each 
end)  on  the  under  side  of 
the  end  post  and  those  at  the  ends  of  the  upper  chord  (four 
in  all)  will  be  21  inches  by  21  inches  by  |  inch.  All  other 
battens  (one  on  that  side  of  each  vertical  post  opposite  to  the 
chord  joint)  will  be  2i  x  15  x  |  inches.  The  bottom  plate  of 
each  joint  forms,  of  course,  a  batten. 

On  the  intermediate  posts,  the  latticing  will  be  single  and 


COMPLETE   DESIG.V  OF  A    RAILWAY  BRIDGE.         4^7 

60°  as  before,  but  the  lattice  bars  will  be  2  inches  by  i\  inch. 
This  latticing  (both  sides)  will  weigh  about  9  pounds  per 
lineal  foot  of  post. 

Floor-beam  Supports. 

The  method  of  suspending  the  floor-beam  from  the  pin  at 
the  lower  extremity  of  brace  2  is  shown  in  Fig.  2,  PI.  XI. 
Two  plates  riveted  to  the  end  stiffeners  of  the  beam  take  the 
5|  pin  with  its  centre  line  six  inches  above  the  upper  flange. 
The  greatest  moving  load  carried  by  the  beam  end  has  been 
already  found  to  be  34,600  pounds.  One-third  of  this  will  be 
added  for  impact,  and  as  the  fixed  load  is  8,000  pounds,  the 
total  load  to  be  resisted  by  the  plate  hangers  becomes  : 

4  X  34.600  + 8,000  =54..30  pounds. 

The  greatest  allowable  load  per  square  inch  in  these  hang- 
ers is  8,000  pounds;  hence  the  required  net  area  is  54,130  -^ 
8,000  =  6.8  square  inches.  These  plates  will  be  taken  12 
inches  wide.  By  deduction  of  the  pin-hole  the  available 
net  width  becomes  12  —  5.375  =  6.625  inches.  One  plate 
12  X  i^g  and  another  12  x  \  inch  gives  the  required  area. 
Rivets  I  inch  in  diameter  will  hold  these  plates  to  the  end 
stiffeners.  The  shearing  resistance  in  this  case  is  less  than 
the  bearing,  and  the  former  for  one  rivet  at  7,500  pounds  per 
square  inch,  is  4,500  pounds.  Hence  the  required  number 
of  rivets  is  54,130  -f-  4,500  —  12  rivets.  In  consequence  of 
the  deflection  of  the  beam  some  of  the  upper  ones  will  be 
subjected  to  slight  tension.  Hence  16  rivets  are  shown.  The 
figure  shows  the  number  and  distribution  of  rivets  and  plates. 
The  vertical  pitch  of  rivets  is  3  inches. 

The  manner  of  attaching  the  floor-beams  to  the  lower  ex- 
tremities of  the  intermediate  posts  is  shown  by  Fig.  14,  PI. 
XI.  a  and  «  are  4  x  4  inch  36  pound  angles  27  inches  long 
riveted  to  the  inner  surfaces  of  the  lo-inch  channels,  as 
shown. 

The  vertical  centre  lines  of  the  rivet  rows  in  the  channels 
27 


4i8 


COMPLETE  DESIGN  OF  A    RAILWAY  BRIDGE. 


are  coincident  with  the  central  Hnes  of  the  latter,  thus  insur- 
ing an  equal  division  of  the  floor-beam  load  between  the 
pin-plates.  The  number  and  distribution  of  the  \  inch  rivets 
in  the  angles  o,  a  will,  of  course,  be  the  same  as  those  in  the 
plate  hangers  of  Fig,  2,  PI.  XL 

Both  these  methods  of  supporting  floor-beams  insure  a 
central  application  to  the  pin,  and  the  latter  insures  addi- 
tional stiffeners  to  the  floor  system  and  entire  structure. 

The  two  lines  of  \  inch  rivets  take  less  than  a  square  inch 
of  section  from  the  two  :o  inch  50 pound  channels.  Hence  the 
remaining  net  section  of  our  nine  square  inches  is  more  than 
sufficient  to  carry  the  total  floor-beam  load  at  all  points. 

Upper  Lateral  System. 

A  wind  pressure  of  150  pounds  per  lineal  foot  will  be  taken 
as  acting  in  the  horizontal  plane  of  the  upper  chord.  The 
panel  wind  load  will  then  be  20.55  ^  150  =  3>o83  pounds. 
Fig.  12  shows  a  half  plan  of  the  upper  lateral  system.  The 
diagonals  are  ties,  and  the  other  members  are  struts. 


Fig.  12. 


The  secant  of  the  inclination  of  T-^  to  a  horizontal  line 
normal  to  the  axis  of  the  bridge  is  1.57,  and  3,083  x  1.57  — 
4,848  pounds. 

The  wind  load  in  the  upper  chord  is  a  fixed  one.  The  line 
ah  is  the  centre  of  the  span. 

The  following  stresses  may  now  be  written,  remembering 
that  as  the  tension  diagonals  will  each  be  adjustable,  5,000 
pounds  must  be  added  for  initial  stress: 


C0MFLE7E   DESIGN'  OF  A    RAILWAY  BRIDGE.         419 
7^4  =:        —     +  5,000  =     5,000  pounds. 

7;  ^   4,848  +     "    =   9,848 

Zj  =    9,696  +       "      =  14,696       " 
Tx  =  14.544  +       "      =  19-544 

Pi  =  3,083  +  3,200  =  6,283  pounds 

/*3  =  6,166  +       "      =  9,366        " 

Pa  =:  9,249  +       "      =  12,449 

P,  =  12,332  +       "      =  15,532 

The  greatest  allowable  stresses  in  the  lareral  systems  may 
be  taken  as  follows  : 


For  tension       ....     14,000  pounds  per  sq.  in 

X) 
7^ 


9,^00  ,     . 

compression.     .     g —  *'         "  (13). 


30,000  r^ 

The  latter  formula  is  for  flat-end  members  of  angle  iron,  as 
those  will  be  used  for  lateral  compression  members.  It  will 
be  observed  that  it  gives  less  value  than  the  formula  (i)  for 
columns  of  the  box  type,  like  those  used  in  the  trusses. 
Under  these  stresses  the  tension  members  become  : 

Ti r  -  ii  o. 

Ts I  -  iJ  o. 

T2 I  —  1}  o. 

Ti I  -  i|  o. 

It  is  not  advisable  to  have  any  tension  member  less  in  sec- 
tional area  than  i  square  inch.  Hence,  T^  and  T^  are  a  little 
larger  than  the  stresses  require. 

All  the  struts  except  P,  will  be  formed  of  3  —  3  x  3  inch 
angles.  A  section  of  this  strut  is  shown  at  Fig.  1 1.  PI.  XI. 
The  two  angles  c  lie  on  the  upper  chord  and  are  riveted  to 
it,  as  shown.  These  two  angles  are  designed  to  carry  all  the 
stress  of  the  strut.  The  only  office  of  the  angle  a  is  to  keep 
the  strut  stiff  in  a  vertical  plane;  it  takes  hold  of  the  lower 
flange  of  the  chord  with  two  rivets,  as  shown  at  c,  Fig.  4,  PI. 


420  COMPLETE   DESIGN  OF  A    RAILWAY  BRIDGE. 

XL  The  strut  is  thus  of  the  same  depth  as  the  chord,  and' 
takes  hold  of  both  those  members  in  such  a  manner  as  to 
give  them  great  rigidity.  At  each  end  of  the  strut  there  is  a 
20  X  12  X  y\  inch  plate,  and  there  is  a  set  of  single  60°  lacing, 
2  X  y^6  inch.  The  radius  of  gyration  of  the  section  of  the 
two  top  angles  about  a  vertical  line  midway  between  the  two, 
is  1.3  inches.  The  length  of  the  strut  is  about  192  inches. 
Hence  Eq.  (13)  gives  5,300  pounds  as  the  greatest  allowable 
stress  per  square  inch.  As  P^  requires  only  the  lightest 
angle  that  ought  to  be  used,  all  these  struts  will  be  made 
alike. 

/s     .     .     .     .      f  3  —  3"  X  2|"  15  pound  angles. 


P. 


As  the  detail  for  these  struts  would  give  some  trouble  at 
the  end  of  the  upper  chord,  P-^  will  be  a  single  6"  x  4'  angle 
with  the  6"  leg  horizontal,  as  shown  at  a.  Fig.  4,  PI.  XI. 

The  tie  7i  is  attached  to  the  upper  chord  by  the  detail 
shown  at  b,  Fig.  11,  PI.  XI.  A  piece  of  6"  by  4",  60  pound 
angle,  12  inches  long,  with  the  6  inch  leg  lying  on  the  cover 
plate  of  the  chord,  carries  two  pieces  of  3  '  by  3",  21  pound 
angles  about  ^\  inches  long,  and  with  edges  parallel  to  the 
axis  of  7"i.  One  end  of  each  of  the  latter  angles  rests  squarely 
against  the  vertical  4-inch  leg  of  the  6"  by  4"  angle.  Six  three- 
quarter  inch  rivets  are  then  passed  through  the  angles  in 
the  manner  shown.  Each  such  rivet  will  resist  about  4,000 
pounds  in  single  shear.  The  tie  T-^  passes  through  the  4-inch 
leg  of  the  heavy  angle  (between  the  3-inch  angles),  and  carries 
a  nut  at  its  end,  which  gives  the  requisite  adjustment. 

Ti  and  all  the  other  lateral  ties  are  held  by  the  same  detail, 
except  that  4  rivets  only  are  needed,  as  shown  at  b\ 

Transverse  Bracing. 

A  skeleton  sketch  of  the  intermediate  transverse  bracing 
for  the  vertical  plane  of  any  two  opposite  posts  is  shown  in 
Fig.  16,  Pi.  XII.     If  the  upper  lateral  system  is  designed  to- 


COMPLE  TE  DESIGN  OF  A    RAIL  IV A  V  BRIDGE.         4^  I 

carry  the  whole  wind  load  to  the  ends  of  the  upper  chord,  as 
has  been  supposed,  the  duty  of  the  intermediate  transverse 
bracing  is  entirely  indeterminate.  As  the  sections  must  be 
determined  in  some  manner,  however,  the  method  of  Art. 
82,  Fig.  I,  will  be  applied. 

The  slight  analytical  superabundance  of  stability  thus 
secured  is  no  more  than  is  required  by  a  rapidly  moving 
load. 

F'  of  Art.  82  will  here  be  taken  as  3,083  pounds  and  F=  o. 
Also,  d  =  1/  ft.  and  a  =  6  ft.  It  will  here  be  assumed  that 
all  the  wind  load  is  applied  in  the  windward  truss.  This  is 
the  usual  assumption  in  practice,  although  the  conditions 
taken  in  Art.  82  are  exactly  true.  Eq.  (i)  of  that  Art.  then 
gives : 

w  =  (3,083  X  27)  -f-  17  =  4,932  pounds. 

As  the  tangent  of  the  inclination  of  7^  to  a  vertical  line  is 
2.833  ^n<^  the  secant,  3.0,  the  stresses  are: 

Z=  4,932  X  3  +  5,000=         19,796  pounds.i I  o. 

P=4,932x  2,833  4- 5,000=  18,974       "       2-3"  x3"24 lb. angles. 

The  sizes  are  based  on  the  same  allowed  working  stresses 
as  for  the  upper  laterals.  The  strut  P  is  shown  at  Fig.  19, 
PL  XII.  The  two  angles  are  held  if  inches  apart  by  separa- 
tors, and  present  a  horizontal  upper  surface.  The  ends  are 
secured  to  a  batten  plate  in  proper  position  on  the  post  and 
in  the  manner  shown.  The  separation  of  the  angles  permits 
the  tie  7"  to  pass  between  them  and  through  the  batten  plate 
and  take  a  nut  inside  the  post.  The  washer  a  is  formed  from 
a  piece  of  3  by  2  inch  angle,  one-half  inch  thick,  with  the  2 
inch  leg  sheared  off  until  the  proper  angle  is  formed.  The  3 
by  2  inch  angle  b  forms  a  check  to  keep  the  angle  washer  a 
in  place. 

The  length  of  the  strut  P  is  192  inches,  while  the  radius  of 
gyration  of  a  3  by  3  inch  angle  about  an  axis  through  its 
centre  of  gravity  and  parallel  to  one  leg  is  0.92  inch.  Hence 
the  allowable  stress  per  sq.  in.  by  Eq.  (13),  is  4,000  pounds. 


422  COMPLETE  DESIGN   OF   A    RAILWAY   BRIDGE. 

Both  ends  of  T  are  held  by  precisely  the  same  detail.  At 
the  upper  end  of  the  post,  however,  the  pin-plates  take  the 
place  of  the  battens  at  the  intermediate  points.  The  rivets 
securing  the  batten  plate  to  the  post  are  seen  to  give  an  ex- 
cess of  resistance. 

Portal  Bracing. 

Fig.  17,  PI.  XII.,  shows  a  skeleton  sketch  of  the  portal 
bracing.  The  sketch  is  taken  in  the  plane  of  the  portal.  The 
strut  Pis  placed  7  ft.  6  in.  from  the  top  of  the  post,  while  the 
length  of  the  latter  is  33.8  ft. 

The  computations  are  made  precisely  as  in  connection  with 
Fig.  16,  PI.  XII.  The  force  acting  at  the  upper  extremity  of 
the  end  post  is  12,332  pounds.  The  tangent  of  the  angle  be- 
tween Zand  the  end  post  is  2.27,  while  the  secant  is  2.48. 
Hence,  if  w  =  12,332  x  33.8  -^  17  =  24,664,  then  : 

T=  24,664  X  2.48  =  61,170  pounds.  .  I  —  6"  X  4"  60  lb.  angle. 
/•=  24,664  X  2.27  =  55,990        "        ..I— 6"  X  6"  75    "       " 

As  shown  by  the  preceding  results,  this  bracing  is  com- 
posed entirely  of  angles.  This  is  done  in  order  to  secure  the 
utmost  stiffness  or  rigidity  in  the  portal. 

Fig.  12,  PI.  XI.,  shows  the  method  of  securing  the  ends  of 
the  members  7"  and  P.  The  upper  extremity  of  7"  is  shown 
with  the  six-inch  leg  of  the  angle  lying  on  the  end  post  at  a. 
In  order  that  the  proper  number  of  three-quarter  inch  rivets 
may  be  brought  into  play,  a  |  inch  plate  lies  underneath  the 
.angle,  as  shown.  The  method  of  securing  the  lower  end  of 
T,  and  each  end  of  P  xs  clearly  shown  at  b.  Three-quarter 
inch  rivets  are  used  for  all  these  connections.  At  the  inter- 
section of  the  two  Ts,  one  is  cut  and  a  firm  joint  is  made  by 
a  centre  plate  a  half-inch  thick,  aided  by  angle  lugs. 

The  length  of  the  strut  P  is  about  192  inches,  and  its  radius 
of  gyration,  1.9  inches.  Hence  Eq.  (13)  gives  the  working 
stress  at  7,000  pounds  per  square  inch.  The  tension  allowed 
in  this  angle  bracing  is  12,000  pounds  per  square  inch. 

An  ornamental  wrought-iron  bracket  may  be  placed  in  the 
angle  between  P  and  the  end  post. 


COMPLETE   DESIGN    OF  A  RAILWAY    BRIDGE.        423 

Lower  Lateral  Bracing. 

The  lower  lateral  bracing  is  shown  in  skeleton  plan  by  Fig. 
'6,  PI.  XII.  It  is  designed  to  resist  a  uniform  fixed  wind  load 
of  150  pounds  per  lineal  foot  in  addition  to  a  uniform  moving 
wind  load  of  300  pounds  per  lineal  foot.  The  fixed  panel 
load,  therefore,  will  be  20.55  x  150=  3,083  pounds,  and  the 
moving  panel  load,  20.55  x  300  =  6,166  pounds.  The  secant 
of  the  angle  between  T^  and  P^  is  1.57.  The  line  ab  is  the 
centre  line  of  the  span.  Remembering  that  there  are  nine 
panels  in  the  lower  lateral  system,  and  that  the  greatest  al- 
lowable tension  is  14,000  pounds,  the  following  stresses  and 
sizes  may  at  once  be  written  from  the  preceding  data: 

7'i  stress 58,164  +  5,000  =  63,164  lbs.  i  —  2"|  round. 

^2      "     44,700+       "      =49700    "     I  -  2"!       " 

^3      "       32,313  +       "      =37,313    "     I  -  I  ^      " 


T^      "     21,000  +      "     =  26,000   "     I  —  I 

7;      "     10,770  +      "     =  15,770   "     I  -  I 


X 


The  initial  stress  is  included  in  the  total  by  the  addition  of 
5,000  pounds,  as  has  been  done  before. 

The  floor-beams  form  the  struts  in  the  lower  lateral  system, 
except  in  the  cases  of  the  end  struts  I\,  hence,  only  the  latter 
need  be  provided.  The  friction  on  the  wall-plate  will  evi- 
dently relieve  P^  of  some  of  its  stress,  but  it  is  uncertain  to 
what  extent ;  hence,  initial  tension  only  will  be  neglected. 
Consequently  : 

Pi  stress 41,620  lbs I  —  6"  x  4  "  50  lb.  angle. 

The  6  inch  leg  of  this  angle  is  horizontal,  and  it  is  riveted 
to  the  top  flanges  of  the  stringers  zvlicre  it  crosses  the  latter.  By 
this  means  the  stringer  ends  are  held  rigidly  in  position,  and 
the  general  stiffness  is  increased. 

The  rods  of  the  lower  lateral  system  will  necessarily  pass 
through  the  webs  of  the  stringers,  and  will  be  secured  to  the 
webs  of  the  floor-beams  (as  closely  as  possible  to  their  upper 
flanges)  by  precisely  the  detail  shown   in  Fig.    19,  PI.  XII. 


424  COMPLETE   DESIGN  OF  A    RAILWAY  BRIDGE. 

The  web-plate  now  takes  the  place  of  the  batten  in  that 
figure. 

The  lower  lateral  7^  takes  hold  of  the  pedestal  by  the  clevis 
and  3  inch  pin  bolt  shown  in  Fig.  lo,  PI.  XI. 

A  9"  X  7''  X  I '  plate  is  riveted  above  and  below  the  f  inch 
base  plate  of  the  pedestal  in  order  to  give  the  proper  bearing 
area.  The  method  of  securing  the  end  of  P,  to  the  pedestal 
is  shown  by  the  same  figure  with  perfect  clearness. 

Expansion  Rollers. 

A  set  of  expansion  rollers  under  one  end  of  each  truss  must 
be  provided.     A  diameter  of  2^  inches  will  be  assumed. 

According  to  Appendix  II.,  the  resistance  per  lineal  inch 
of  a  roller  is: 


4 


^|/.^^'  =  |^|/4^(for£  =  £> 


Since  all  metal  is  wrought-iron  E  =  E' .  The  greatest  al- 
lowable intensity  of  pressure  on  the  roller  will  be  taken  at 
12,000  pounds  per  square  inch,  or  «;  =  12,000.  Also  R  =  1.47 
and  E  —  26,000,000.  Hence  the  allowable  load  per  lineal 
inch,  by  the  above  formula,  is  1,050  pounds.  The  maximum 
vertical  component  of  the  end-post  stress  is  172,820  pounds^ 
Hence  the  number  of  lineal  inches  of  roller  bearing  required 
is  172,820  -=-  1,050  =  164  inches.  The  set  of  rollers  shown  by 
Fig.  II,  PI.  XII.,  gives  very  closely  the  required  amount. 
The  clear  space  between  each  adjacent  pair  of  rollers  is  ^  inch. 
The  ends  of  each  roller  are  turned  down  to  |  inch  and  pass 
through  a  2i  x  |  inch  wrought-iron  strap  on  the  outside  of 
which  each  roller  end  takes  a  nut.  The  rollers  are  thus  held 
rigidly  in  their  proper  relative  positions.  A  i  x  ^  inch  collar 
is  turned  down  at  the  centre  of  each  roller  to  take  the  i  x  y'^ 
inch  shoulder  which  is  shown  in  the  wall  plate,  and  by  means 
of  which  all  lateral  motion  of  the  rollers  is  prevented. 

Wall  Plates. 
The  mean  pressure  per  square  inch  on  the  total  surface  of 


COMPLETE   DESIGN  OF  A    RAILWAY   BRIDGE.         4-5 

the  wall  plate  should  not  exceed  200  pounds.  The  total  area 
of  wall-plate  surface  shown  in  Fig.  12,  PL  XII.,  is  30  x  30  = 
900;  hence  the  total  allowable  weight  is  900  x  200  =  180,000 
pounds.  The  maximum  total  vertical  pressure  of  172,820 
pounds  is  thus  provided  for. 

The  plan  of  the  wall  plate  at  the  roller  end  is  shown  in  the 
figure ;  the  upper  elevation  also  belongs  to  it.  At  the  fixed 
end  either  the  pedestal  or  the  masonry  must  be  sufficiently 
high  to  fill  the  roller  space.  Both  those  alterations,  however, 
are  unadvisable  for  obvious  reasons.  It  is  better  to  fill  the 
roller  space  with  the  wall  plate.  The  lower  elevation  of  Fig. 
12,  PI.  XII.,  shows  the  arrangement  to  be  adopted.  The 
same  |  inch  thick  wall  plate  is  to  be  taken,  4  —  3i  x  3^  x  | 
inch  angles  then  run  in  the  direction  of  the  rollers  across  the 
■entire  plate.  At  right  angles  to  these  /\  lines  of  3  x  3  x  4 
inch  angles  are  placed.  The  latter  are  cut  to  fit  in  between 
the  former  and  will,  of  course,  require  filling  strips  under- 
neath. The  top  of  this  gridiron  arrangement  is  then  planed 
ofl  until  the  proper  height  of  wall  plate  is  reached.  The 
rectangular  spaces  thus  formed  are  then  filled  with  Portland 
cement  rammed  hard  and  flush  with  the  planed  upper  sur- 
faces of  the  angles.  A  solid  wall  plate  is  thus  formed  with 
the  interior  surfaces  completely  protected  against  corrosion. 

At  diagonally  opposite  corners  are  seen  the  holes  for  the 
i|  inch  anchor  bolts. 

Wind  Pressure  on  Chords  and  End  Posts. 

The  effect  of  the  wind  load  on  both  upper  and  lower  chords 
has  been  shown  in  detail  in  Art.  81  ;  the  principles  there 
established  remain  to  be  applied  here.  The  chord  stresses  in 
the  lateral  trusses  are  the  same  in  kind  as  those  produced  by 
the  vertical  loading  in  one  lower  chord  and  one  upper  chord. 
Just  to  what  extent  these  wind  stresses  may  be  allowed  to 
exist  without  necessitating  any  increased  chord  section  is  a 
matter  of  experience  only ;  but  as  the  greatest  wind  stresses 
and  those  due  to  the  vertical  loading  so  rarely  combine  in 
jnost  localities  that  with  the  working  stresses  specified  in  this 


426  COMPLETE   DESIGN   OE  A    RAILWAY  BRIDGE. 

case    the  7vind  load  may  he  allowed  to  reach   \tJis  the  value 

of  the  greatest  vertical  loading  without  requiring  any  increase 

in  chord  section,   in    all    localities  not    ordinarily  subject    to 

cyclones  or  tornadoes. 

In  the  present  instance  the  total  fixed  and  moving  vertical 

load  is  equivalent  to  about  2,040  pounds  per  lineal  foot  of 

each  truss.     The  total  wind  load  in  the  lower  chord  is  450- 

pounds  per   lineal   foot,  and    its  depth   of'  truss   is  only    17 

feet.     If    reduced  to  the  same  truss  depth  ao  that    for   the 

27 
vertical  load,  /.  r.,  27  feet,  it  would  be  450  x  —  =  720  pounds. 

Again,  the  overturning  effect  of  the  wind  on  the  train  (dis- 
cussed at  the  close  of  Art.  81)  throws   on  the  leeward  truss 

the  additional  weight  of  ^ =  140  pounds  per  lineal  foot. 

It  is  assumed  that  the  centre  of  wind  pressure  on  the  train  is 
8  feet  above  the  end  supports  of  the  floor-beam.  The  total 
wind  effect  on  the  loading  of  the  leeward  truss  is  then 
720  +  140  =  860  pounds  per  lineal  foot,  or  a  little  in  ex- 
cess of  four-tenths  the  vertical  loading.  As  three-eighths 
the  vertical  loading  is  765  pounds  per  lineal  foot,  the  chord 
sections  should  be  increased  for  95  pounds  per  lineal  foot. 
As  the  increase  in  area,  however,  would  be  but  one-sixth 
of  an  inch  for  one  chord,  and  as  the*  transverse  bracing 
will  slightly  relieve  the  leeward  truss,  no  change  will  be 
made. 

If  the  lower  chord  needs  nc  revision,  the  upper  need  not 
be  considered. 

The  total  pressure  of  wind  against  the  upper  extremities  of 
the  end  posts,  and,  hence,  against  their  lower  extremities 
also,  has  already  been  seen  to  be  12,332  pounds.  If  this  is 
assumed  to  be  equally  divided  between  the  end~post  feet, 
each  of  the  latter  will  carry  6,166  pounds.  Each  end  of 
P(in  the  portal.  Fig.  17,  PI.  XII.)  is  26.3  feet  from  the  end- 
post  foot.  Hence  at  the  former  point  the  end  post  suffers 
the  bending  moment  6,166  x  12  x  26.3  =  1,945,990  in.  lbs. 
The  moment  of  inertia  of  the  end-post  section  about  the 
neutral  axis  normal  to  the  cover  plate  is  1,886,  and  since  the 


COMPLETE   DESIGN  OF  A    RAILWAY  BRIDGE.  4-7 

half  total   width  of   the  chord  is   12.5   inches,  the  stress  per 
square  inch  in  the  extreme  fibres  of  the  5x3  angles  is 

K'  =  1,945,990  X  12.5  -f-  1,886  =  i2,go4  pounds  per  sq.  in. 

The  direct  post  stress  will  be  about  7,200  more,  or  a  total 
of  20,104  pounds  per  sq.  in.,  whereas  15,000  should  not  be  ex- 
ceeded. 

\{  K  —  limit  of  compressive  bending  stress  per  square  inch, 
which  must  not  be  exceeded  ;  M  =■  bending  moment  in  inch 
pounds  ;  /  the  moment  of  inertia  of  the  total  post  section 
about  a  neutral  axis  normal  to  the  cover-plate  ;  2d  =  total 
width  of  end  post ;  P  =  total  direct  stress  of  compression, 
due  to  vertical  loading  and  overturning  action  of  the  wind 
against  the  train  (/  of  Art.  81  is  its  panel  value),  and  A  = 
total  area  of  section  ;  then  the  sectional  area  must  be  increased 
until  the  following  equation  holds  true  : 

'          Md       P  ,     , 

^=^^A ^^4). 

It  has  been  seen  above  that  the  wind  effect  in  the  leeward 
truss  is  140  lbs.  per  lin.  ft.,  or  20.55  ^  HO  =  2,877  lbs.  per 
panel.  Hence,  4  x  2,877  ^  ^-26  =  14,500  lbs.  is  that  part 
of  P  due  to  the  wind  ;  or : 

P—  14,500  +  217,750  =  232,250  lbs. 
Also,  2^  =  25  ;  ov  d  =  12.5  inches ; 
And,  M  =  1,945,  990  /;/.  lbs. 

If  2  —  3"  X  3"  25  lb.  angles  are  riveted  on  the  outside  of 
each  side  plate  of  each  post  in  the  manner  shown  in  Fig.  13, 
the  centres  of  gravity  of  those  angles  will  be  8.3  inches  from 
the  neutral  axis  about  which  /  is  taken,  and  the  moment  of 
inertia  of  each  angle  section  about  a  parallel  axis  through  its 
own  centre  of  gravity  is  2.25;  hence: 

/=  1,886  +  4  X   2.5   X  (8.3)2  +  4  X   2.25  =  2,585. 

Finally  : 

A  =  32.3  +  4  X  2.5  =  42.3  sq.  ins. 


428  COMPLETE   DESIGN   OF  A    RAILWAY  BRLDGE. 

These  quantities  placed  in  Eq.  (14)  give: 

K  =  9,412  +  5,500  =  14,912  lbs.  per  sq.  in. 

which  shows  that  the  desired  section  is  obtained. 

Fig.  13  shows  an  elevation  of  parts  of  the  end  post,  which 
is  supposed  to  be  intersected  by  the  portal  strut  at  ab.  cc' ,  and 
dd'  are  the  3"  x  3"  25  lb.  angles.  ^  is  9  ft.  below  ab  and  c' 
2  ft.  9  in.  above  it.  dd'  is  half  the  length  of  cc'.  Below  c 
and  above  c' ,  the  preceding  figures  show  that  no  increase  of 
section  is  needed. 

If  much  increase  is   needed,  unequal   legged   angles  with 


id' 


Fig.  13. 

the  larger  legs  normal  to  the  side  plates  can  be  most  advan- 
tageously used. 

It  will  ordinarily  be  suf^ciently  accurate  to  increase  the 
K' 


section  in  the  ratio  of 


K 


These  computations  show  what  an  important  factor  the 
wind  load  may  be  in  a  country  subject  to  tornadoes  and  cy- 
clones. In  such  exposed  localities  the  chord-wind  stresses 
should  not  be  allowed  to  exceed  25  per  cent,  of  those  due  to 
the  vertical  loading  without  providing  correspondingly  in- 
creased sections. 

The  wind  effect  on  the  stringers  mentioned  in  Art.  81  is 
such  a  small  per  centage  of  the  vertical  load  that  it  need  not 
be  considered. 

Conclusion. 

It  is  not  necessary  here  to  produce  in  detail  the  complete 
list  of  weights  of  all  the  parts,  although  this  must  invariably 


COMPLETE   DESIGN  OF  A    RAILWAY  BRIDGE.        429 

6e  done  in  practice.  If  the  estimated  weight  comes  out 
greater  than  the  assumed,  a  revision  of  the  computations 
must  be  made  with  a  sufficiently  increased  fixed  weight  to 
exceed,  at  least  by  a  little,  the  estimated  weight.  If,  on  the 
other  hand,  the  estimated  weight  is  considerably  less  than 
the  assumed,  the  latter  may  be  reduced  in  a  recomputation, 
in  order  to  reach  a  proper  degree  of  economy. 


APPENDIX  I. 

THE  THEOREM  OF  THREE  MOMENTS. 

Art.  I. — The  object  of  this  theorem  is  the  determination 
of  the  relation  existing  between  the  bending  moments  which 
are  found  in  any  continuous  beam  at  any  three  adjacent 
points  of  support.  In  the  most  general  case  to  which  the 
theorem  applies,  the  section  of  the  beam  is  supposed  to  be 
variable,  the  points  of  support  are  not  supposed  to  be  in  the 
same  level,  and  at  any  point,  or  all  points,  of  support  there 
may  be  constraint  applied  to  the  beam,  external  to  the  load 
which  it  is  to  carry ;  or,  what  is  equivalent  to  the  last  condi- 
tion, the  beam  may  not  be  straight  at  any  point  of  support 
before  flexure  takes  place. 

Before  establishing  the  theorem  itself,  some  preliminary 
matters  must  receive  attention. 

In  Fig.  I,  let  ABC  represent  the  centre  line  of  any  bent 
beam;  AF,  a  vertical  line  through  A  ;   CF,  a  horizontal  line 


Fig   I. 


through  C,  while  A  is  the  section  of  the  beam  at  which  the 
deflection  (vertical  or  horizontal)  in  reference  to  C,  the  bend- 
ing moment,  the  shearing  stress,  etc.,  are  to  be  determined. 

431 


43' 


APPENDIX  I. 


As  shown  in  the  figure,  let  x  be  the  horizontal  co-ordinate 
measured  from  A,  and  y  the  vertical  one  measured  from  the 
same  point  ;  then  let  z  be  the  horizontal  distance  from  the 
same  point  to  the  point  of  application  of  any  external  ver- 
tical force  P.  To  complete  the  notation,  let  D  be  the  deflec- 
tion desired  ;  M^,  the  moment  of  the  external  forces  about  A  ; 
S,  the  shear  at  A  ;  P' ,  the  strain  (extension  or  compression) 
per  unit  of  length  of  a  fibre  parallel  to  the  neutral  surface 
and  situated  at  a  normal  distance  of  unity  from  it ;  /,  the 
general  expression  of  the  moment  of  inertia  of  a  normal  cross- 
section  of  the  beam,  taken  in  reference  to  the  neutral  axis  of 
that  section  ;  E  the  coefficient  of  elasticity  for  the  material 
of  the  beam;  and  M  the  moment  of  the  external  forces  for 
any  section,  as  B. 

Again,  let  A  be  an  indefinitely  small  portion  of  any  normal 
cross-section  of  the  beam,  and  let  y  be  an  ordinate  normal 
to  the  neutral  axis  of  the  same  section.  By  the  "  common 
theory  "  of  flexure,  the  intensity  of  stress  at  the  distance  y 
from  the  neutral  surface  is  {y'P'E).  Consequently  the  stress 
developed  in  the  portion  A,  of  the  section,  is  EP'y  A,  and  the 
resisting  moment  of  that  stress  is  EP'y'^A. 

The  resisting  moment  of  the  whole  section  will  therefore 
be  found  by  taking  the  sum  of  all  such  moments  for  its 
whole  area. 

Hence : 

M=EP':Sy"-A  =  EP'L 

Hence,  also, 

EI 

If  n  represents  an  indefinitely  short  portion  of  the  neutral 
surface,  the  strain  for  such  a  length  of  fibre  at  unit's  distance 
from  that  surface  will  be  nP'. 

If  the  beam  were  originally  straight  and  horizontal,  n  would 
be  equal  to  dx. 

P'  being  supposed  small,  the  effect  of  the  strain  7iP'  at  any 
section,  B,  is  to  cause  the  end  K,  of  the  tangent  BK,  to  move 
vertically  through  the  distance  nP'x. 


THEOREM   OF  THREE  MOMENTS.  433 

If  BK  2iX\A  BR  (taken  equal)  are  the  positions  of  the  tan- 
gents before  and  after  flexure,  7iP' x  will  be  the  vertical  dis- 
tance between  TsTand  R. 

By  precisely  the  same  kinematical  principle,  the  expression 
nP'v  will  be  the  horizontal  movement  of  A  in  reference  to  B. 

Let  '2nP'x  and  ^nP'y  represent  summations  extending 
from  A  to  C,  then  will  those  expressions  be  the  vertical  and 
horizontal  deflections,  respectively,  of  A  in  reference  to  C. 
It  is  evident  that  these  operations  are  perfectly  general,  and 
that  X  and  y  may  be  taken  in  any  direction  whatever. 

The  following  general,  but,  strictly,  approximate  equations, 
relating  to  the  subject  of  flexure,  may  now  be  written  : 

S    =:2P (i). 

M^  =  2Ps (2). 

2nP'=2n^ (4). 

D  =2nP'x=^"^       ....     (5). 

A  =  2.P>=^^^      ....     (6). 

Z>ft  represents  horizontal  deflection. 

Art.  2. — Some  elementary  but  general  considerations  in 
reference  to  that  portion  of  a  continuous  beam  included  be- 
tween two  adjacent  points  of  support  must  next  be  noticed. 

If  a  beam  is  simply  supported  at  each  end,  the  reactions 

are  found' by  dividing  the  applied   loads   according  to  the 

simple  principle  of  the  lever.      If,  however,  either  or  both 

ends  are  not  simply  supported,  the  reaction,  in  general,  is 

greater  at  one  end  and  less  at  the  other,  than  would  be  found 
28 


434 


APPENDIX  I. 


by  the  law  of  the  lever ;  a  portion  of  the  reaction  at  one  end 
is,  as  it  were,  transferred  to  the  other.  This  transference  can 
only  be  accomplished  by  the  application  of  a  couple  to  the 
beam,  the  forces  of  the  couple  being  applied  at  the  two  adja- 
cent points  of  support  ;  the  span,  consequently,  will  be  the 
lever  arm  of  the  couple.  The  existence  of  equilibrium  re- 
quires the  application  to  the  beam  of  an  equal  and  opposite 
couple.  It  is  only  necessary,  however,  to  consider,  in  connec- 
tion with  the  span  AB,  the  one  shown  in  Fig.  2.  Further, 
from  what  has  immediately  preceded,  it  appears  that  the 
force  of  this  couple  is  equal  to  the  difference  between  the 
actual  reaction  at  either  point  of  support  and  that  found  by 
the  law  of  the  lever.  The  bending  caused  by  this  couple  will 
evidently  be  of  an  opposite  kind  to  that  existing  in  a  beam 
simply  supported  at  each  end. 

These  results  are  represented  graphically  in  Fig.  2.  A  and 
B  are  points  of  support,  and  AB  is  the  beam;  AR  and  BR' 
are  the  reactions  according  to  the  law  of  the  lever;  RF=i 


Fig.  2. 


R'F  is  the  force  of  the  applied  couple;  consequently  ^7^  = 
AR  +  RF  and  BF=  BR'  ~  {R'F=  RF)  are  the  reactions 
after  the  couple  is  applied.  As  is  well  known,  lines  parallel 
to  CK,  drawn  in  the  triangle  A  CB,  represent  the  bending  mo- 
ments at  the  various  sections  of  the  beam,  when  the  reac- 
tions are  AR  and  BR'.    Finally,  vertical  lines  parallel  to  AG, 


THEOREM  OF   THREE   MOMENTS.  4.- 

in  the  triangle  QHG,  will  represent  the  bending  moments 
caused  by  the  force  R'F. 

In  the  general  case  there  may  also  be  applied  to  the  beam 
two  equal  and  opposite  couples,  having  axes  passing  through 
A  and  B  respectively.  The  effect  of  such  couples  will  be 
nothing  so  far  as  the  reactions  are  concerned,  but  they  will 
cause  uniform  bending  between  A  and  B.  This  uniform  or 
constant  moment  may  be  represented  by  vertical  lines  drawn 
parallel  to  AH  or  LN  (equal  to  each  other)  between  the  lines 
AB  and  HQ.  The  resultant  moments  to  which  the  various 
sections  of  the  beam  are  subjected  will  then  be  represented 
by  the  algebraic  sum  of  the  three  vertical  ordinates  included 
between  the  lines  ACB  and  GQ.  Let  that  resultant  be 
called  iM. 

Let  the  moment  GA  be  called  Ma,  and  the  moment  BQ  — 
LN  —  HA,  Ml,.  Also  designate  the  moment  caused  by  the 
load  P,  shown  by  lines  parallel  to  CK  in  ACB,  by  J/j.  Then 
let  X  be  any  horizontal  distance  measured  from  A  towards  B ; 
./the  horizontal  distance  AB\  and  s  the  distance  of  the  point 
of  application,  K,  of  the  force  /*  from  A.  With  this  notation 
there  can  be  at  once  written : 

M=Ma(^-^)+M,{^-^  +  M,.     .     .     (7). 

Eq.  (7)  is  simply  the  general  form  of  Eq.  (2). 

It  is  to  be  noticed  that  Fig.  2  does  not  show  all  the  mo- 
ments Ma,  Mb  and  M^  to  be  of  the  same  sign,  but,  for  con- 
venience, they  are  so  written  in  Eq.  (7). 

Art.  3. — The  formula  which  represents  the  theorem  of 
three  moments  can  now  be  written  without  difficulty.  The 
method  to  be  followed  involves  the  improvements  added  by 
Prof.  H.  T.  Eddy,  and  is  the  same  as  that  given  by  him  in  the 
"American  Journal  of  Mathematics,"  Vol.  I.,  No.  i. 

Fig.  3  shows  a  portion  of  a  continuous  beam,  including 
two  spans  and  three  points  of  support.  The  deflections  will 
be  supposed  measured  from  the  horizontal  line  NQ.  The 
spans  are  represented  by  4  and  4 ;  the  vertical  distances  of 


436 


APPENDIX  I. 


NQ  from  the  points  of  support  by  Ca,  Cj,  and  C(, ;  the  moments 
at  the  same  points  by  Ma,  M^  and  M^,  while  the  letters  5  and 
R  represent  shears  and  reactions  respectively. 


2V 


2^ 


# 


o 


R 


Fig.  3. 


In  order  to  make  the  case  general,  it  will  be  supposed  that 
the  beam  is  curved  in  a  vertical  plane,  and  has  an  elbow  at  b^ 
before  flexure,  and  that,  at  that  point  of  support,  the  tangent 
of  its  inclination  to  a  horizontal  line,  toward  the  span  4  is  /, 
while  /'  represents  the  tangent  on  the  other  side  of  the  same 
point  of  support ;  also  let  d  and  d'  be  the  vertical  distances, 
before  bending  takes  place,  of  the  points  «  and  r,  respectively, 
below  the  tangents  at  the  point  b. 

A  portion  of  the  difference  between  r„  and  r^  is  due  to  the 
original  inclination,  whose  tangent  is  /,  and  the  original  lack 
of  straightness,  and  is  not  caused  by  the  bending ;  that  por- 
tion which  is  due  to  the  bending,  however,  is,  remembering 
Eq.  (5) : 

D  =  Ca  —  c^  —  IJ  -  d=  2^  —^jr- 
By  the  aid  of  Eq.  (7)  this  equation  may  be  written : 


£{(^a~(^b-^a^-^)  =  ^l 


L) 


M„ 


I 
M, 


+ 


(8). 


In  this  equation,  it  is  to  be  remembered,  both  x  and  z  (in- 
volved in  J/i)  are  measured  from  support  a  toward  support 
b.     Now  let  a  similar  equation  be  written  for  the  span  4,  in 


THEOREM  OF   THREE  MOMENTS.  437 

which  the  variables  x  and  z  will  be  measured  from  c  toward 
b.     There  will  then  result : 

E(c,  -c-  IJ'  -  «")  =  i'^  [  ]  i/,  (^)  +  Ah  (7)  + 

^,}f]     .     .     .     (9). 

When  the  general  sign  of  summation  is  displaced  by  the 
integral  sign,  71  becomes  the  differential  of  the  axis  of  the 
beam,  or  ds.  But  ds  may  be  represented  by  udx,  u  being  such 
a  function  of  x  as  becomes  unity  if  the  axis  of  the  beam  is 
originally  straight  and  parallel  to  the  axis  of  x.  The  Eqs. 
(8)  and  (9)  may  then  be  reduced  to  simpler  forms  by  the  fol- 
lowing methods  : 

In  Eq.  (8)  put 


(10). 


a  fl  —  x\  xn  __  I     /*"  u  {la  —  x)  xdx  _ 

fa     r"    U{l^-X)dx 

Also, 
Also, 

'^r  U  (4  -X)dx  =  '-^^f\la  -X)dx  = 


^a  ^a  ^^a  'a 
2 

.      .      (12). 

In  the  same  manner: 

^.a  x^n       I     /*"  iix^dx 
'laI~Ub          I 

^d   P'^  uxdx 

~    U  b           I 

.      .      (13). 

Also, 

Xa     /*"  Uxdx 
laJb          I 

=  '''''.''  f\xdx     . 

.      (14). 

438  APPENDIX  I. 

And, 


h^  f\xdx  =  ^^  f\dx  =  ^^^^^^"^^    .    .     (15). 
Again,  in  the  same  manner: 

^l — Y^  =hathn^  M^^A  X     .     .     .     (16). 
Using  Eqs.  (10)  to  (16),  Eq.  (8)  may  be  written: 

E{c^-C^-lat  -  d)   =^  {Ma  Ua  4  X^  +  M^  11^  la^a  )  + 

Uxaha^\MxA  X       .       .       .       (17). 

Proceeding  in  precisely  the  same  manner  with  the  span  4, 
Eq.  (9)  becomes : 

£{cc-ci,-  /,f-d')  =  ^  {M^u^ i^x,  +  M^,u;i;x; )  + 

ti,,i,,^\M,xAx     .     .     .     (18). 

The, quantities  Xg  and  x^.  are  to  be  determined  by  applying 
Eq.  (10)  to  the  span  indicated  by  the  subscript ;  while  Ua,  ia, 
Uc  and  Zc  are  to  be  determined  by  using  Eqs.  (11)  and  (12)  in 
the  same  way.  Similar  observations  apply  to  2ia,  4  ,  x„,  n^y 
ij  and  x^,  taken  in  connection  with  Eqs.  (13),  (14)  and  (15). 

If  /  is  not  a  continuous  function  of  x,  the  various  integra- 
tions of  Eqs.  (10),  (i  i),  (13),  and  (14)  must  give  place  to  sum- 
mations (-2")  taken  between  the  proper  limits. 

Dividing  Eqs.  (17)  and  (18)  by  4  and  4.  respectively,  and 
adding  the  results : 

\     la  4  4       4  /         4 


I  * 

McUcicXc+  Mi,Uci^'Xc)     ....     (19). 


in  which  T=  t  +  f 


THEOREM  OF   THREE   MOMEA'TS.  439 

Eq.  (19)  is  the  most  general  form  of  the  theorem  of  three 
moments  if  E,  the  coefificient  of  elasticity,  is  a  constant  quan- 
tity. Indeed,  that  equation  expresses,  as  it  stands,  the  "  the- 
orem "  for  a  variable  coefficient  of  elasticity  if  {ie)  be  written 
instead  of  i  \  e  representing  a  quantity  determined  in  a  man- 
ner exactly  similar  to  that  used  in  connection  with  the  quan- 
tity i. 

In  the  ordinary  case  of  an  engineer's  experience,  T=  o, 
/i  =  d'=:  o,  /=  co)istant,  11  =  ii^  =  ii^  =  etc.  =  c  =i  secant  of  the 
inclination  for   ivhicJi   t  —  —  t' is  the   tangent;  consequently 

•_•'_•_•'_•    _■    —I 

^a  —  ^a    —  ^c  —  ^  c  —  ^ ,a  —  ^,c  —  ~J' 

From  Eq.  (10), 

2/„  2/. 


Xa-      ^     ,  -c-      ^ 


From  Eq.  (13), 

«  ~    6  '  '  ~    6   ' 

The  summation  ^M^xAx  can  be  readily  made  by  referring 
to  Fig.  2. 

The  moment  represented  by  CK  in  that  figure  is, 

consequently  the  moment  at  any  point  between  A  and  K,  due 
to  F,  is, 

Between  K  and  B, 


Using  these  quantities  for  the  span  4 


440  APPENDIX  I. 

^lul/.xJx  =JlM^xdx  +f'"M,xdx  =  \P{la^  -  ^)z. 

For  the  span  4,  the  subscript  a  is  to  be  changed  to  c. 
Introducing  all  these  quantities,   E(\.  (19)  becomes,  after 
providing  for  any  number  of  weights,  P: 

C      \        1(1  if,        / 

}-iP{l^-z')z^^kp{Q-£^)z.     .     .     .     (20). 

Eq.  (20),  with  <:'  equal  to  unity,  is  the  form  in  which  the 
theorem  of  three  moments  is  usually  given  ;  with  c'  equal  to 
unity  or  not,  //  applies  only  to  a  beam  whicJi  is  straight  before 
flexure,  since  7"=  t-Vt'=o  —  d  —  d'. 

If  such  a  beam   rests  on  the  supports  a,  b,  and  e,  before 

bending  takes  place,  ^'~ — -  = ^— ^ — -,  and    the  first  mem- 

ta  'c 

ber  of  Eq.  (20)  becomes  zero. 

If,  in  the  general  case  to  which  Eq.  (19)  applies,  the  deflec- 
tions ea,  Ci,,  and  e^  belong  to  the  beam  in  a  position  of  no 
bending,  the  first  member  of  that  equation  disappears,  since 
it  is  the  sum  of  the  deflections  due  to  bending  only,  for  the 
spans  4  and  4)  divided  by  those  spans,  and  each  of  those 
quantities  is  zero  by  the  equation  immediately  preceding,  Eq. 
(8).  Also,  if  the  beam  or  truss  belonging  to  each  span  is 
straight  between  the  points  of  support  {such  points  being  sup- 
i>o.sed  in  the  same  level  or  not),  tia=  i^n  =  ^ha  —  constant,  and 
Uf.—  Uc  =  Uic  =  another  constant.  If,  finally,  /be  again  taken 
as  constant,  Xa  and  x^,  as  well  as  ^M^xAx,  will  have  the  values 
found  above. 

From  these  considerations  it  at  once  follows  that  the 
second  member  of  Eq.  (20),  put  equal  to  zero,  expresses  the 
theorem  of  three  moments  for  a  beam  or  truss  straight  be- 
tween points  of  support,  when  those  points  are  not  in  the 
same  level,  but  when  they  belong  to  a  configuration  of  no 
bending  in  the  beam.  Such  an  equation,  however,  does  not 
belong  to  a  beam  not  straight  between  points  of  support. 


THEOREM  OF    THREE  MOMENTS. 


441 


The  shear  at  either  end  of  any  span,  as  l„,  is  next  to  be 
found,  and  it  can  be  at  once  written  by  referring  to  the  ob- 
servations made  in  connection  with  Fig.  2.  It  was  there  seen 
that  the  reaction  found  by  the  simple  law  of  the  lever  is  to 
be  increased  or  decreased  for  the  continuous  beam,  by  an 
amount  found  by  dividing  the  difference  of  the  moments  at 
the  extremities  of  any  span  by  the  span  itself.  Referring, 
therefore,  to  Fig.  3,  for  the  shears  5,  there  may  at  once  be 
written  : 

5^  =  i/>^L^_^!^^LZ_^     .     .     .     (21). 


5,'=:SPf  +^^^-^*        ....      (22). 
*a  'a 

s,  =  :ep^+'^^^-^    ....   (23). 

The  negative  sign  is  put  before  the  fraction  — ^ ~ ,  in 

'a 

Eq.  (21),  because  in  Fig.  2  the  moments  J/^  and  J/^  are  rep- 
resented opposite  in  sign  to  that  caused  by  P,  while  in  Eq. 
(7)  the  three  moments  are  given  the  same  sign,  as  has  already 
been  noticed. 

Eqs.  (21)  to  (24)  are  so  written  as  to  make  an  upward  re- 
action positive,  and  they  may,  perhaps,  be  more  simply  found 
by  taking  moments  about  either  end  of  a  span.  For  example, 
taking  moments  about  the  right  end  of  4: 

SJa  -  ^P{la  -  =)  +  Ma  =  M^. 

From  this,  Eq.  (21)  at  once  results.  Again,  moments  about 
the  ieft  end  of  the  sam.e  span  give  : 


442  APPEiVDIX  I. 

This  equation  gives  Eq.  (22),  and  the  same  process  will 
give  the  others. 

If  the  loading  over  the  different  spans  is  of  uniform  inten- 
sity, then,  in  general,  P  —  wdz;  w  being  the  intensity.  Con- 
sequently : 

:SP(/2  -  ^)  ^  =  /w(/2  -  z')zdz  =  w  -. 
^  o  4 

I    " 
In  all  equations,  therefore,  for  —  '2  P{la  —  z^)  z  there  is  to 

/^  I      ^ 

be  placed  the  term  w^-- ;  and  for  —  ^ P{l^f  —  ^)  z,  the  term 
4  4 

43 
Wg—  .     The  letters  a  and  c  mean,  of  course,  that  reference  is 

4 

made  to  the  spans  4  and  4. 

From  Fig.  3,  there  may  at  once  be  written : 

R      =Sa     +   S, (25). 

R'  =S,   +  S, (26). 

R"  =67   +  5, (27).. 

ctc.=  etc.  +  etc. 


APPENDIX    II. 

THE   RESISTANCE   OF   SOLID    METALLIC   ROLLERS. 

An  approximate  expression  for  the  resistance  of  a  rollei 
may  easily  be  written,  and  although  the  approximation  may 
be  considered  a  loose  one,  it  furnishes  an  excellent  basis  for 
an  accurate  empirical  formula. 

The  following  investigation  contains  the  improvements  by 
Prof.  J.  B.  Johnson  and  Prof.  H.  T.  Eddy  on  the  method 
originally  given  by  the  author. 

The  roller  will  be  assumed 
to  be  composed  of  indefi- 
nitely thin  vertical  slices  par- 
allel to  its  axis.  It  will  also 
be  assumed  that  the  layers 
or  slices  act  independently 
of  each  other. 

Let  E  be  the  coefficient 
of  elasticity  of  the  metal  over 
the  roller. 

Let  E  be  the  coefificient 
of  elasticity  of  the  metal 
of  the  roller.  y\g.  i. 

Let  R  be  the  radius  of  the 
roller  and  R'  the  thickness  of  the  metal  above  it. 


Let  w  =  intensity  of  pressure  at  A. 
"    /   =         "  "  any  other  point. 

"  /•  =  total  weight  which  the  roller  sustains  per  unit  of 

length. 
"   X  be  measured  horizontally  from  A  as  the  origin. 
"   ^  =  AC. 
"   -f    =  DC. 


444  APPENDIX  II. 

From  Fig.  I  : 

h  h 

h  h 

,'.  d  =  AC  =  AB  +  BC  =  w  (^  +  I') ; 
And 


(I). 


A'C'  =  A'B'  +  B'C:=^p(^  +  ^)l.     .    .    .    (2). 
Dividing  Eq.  (2)  by  Eq.  (i): 


p  =  A'C  -^ 


But 


p-^r  'pdx  =  '^  r '  A'C'dx. 


If  the  curve  DAH  be  assumed  to  be  a  parabola,  as  may  be 
done  without  essential  error,  there  will  result : 

/''  A'Cdx  =  -ed. 
e  3 

Hence : 

P=^we (3). 

But: 

e  =  V  2J^d  -  d-  =  V  2^  nearly. 

By  inserting  the  value  of  d  from   Eq.  (i)  in  the  value  of  e, 
just  determined,  then  placing  the  result  in  Eq.  (7) : 

P=t^lJiJf4) (4). 


THE   RESISTANCE    OF  SOLID    METALLIC  ROLLERS.      445 


P=^  R 


V 


2lif 


,E  +  E' 
EE' 


(5). 


The  preceding  expressions  are  for  one  unit  of  length.     If 
the  length  of  the  roller  is  /,  its  total  resistance  is 


i/' 


P  =  P'  =  \l\/   2^R{^^§)     ...     (6). 


Or  if  ^  =  K 


/ 


^'=f^V-^^^' 


.     (7). 


In  ordinary  bridge  practice  Eq.  (7)  is  sufficiently  near  for 
all  cases. 

A  simple  expression  for  conical  rollers  may  be  obtained  by 
using  Eqs.  (4)  or  (5). 

As  shown  in  Fig.  2,  let  s  be  the  distance,  parallel  to  the 
.axis,  of  any  section  from  the  apex  of  the  cone;  then  consider 

J, .      -_-^, 


Fig.  2. 


■a  portion  of  the  conical  roller  whose  length  is  dz.  Let  Rx  be 
the  radius  of  the  base.  The  radius  of  the  section  under  con- 
-sideration  will  then  be 


R^yR.: 


44^  APPENDIX  II. 

and  the  weight  it  will  sustain,  if  R^  =  R  ; 


R,      /         E  +  E' 
dP'  =  y  A/  2w^  —EE^'  ^^^' 

Hence  : 


^■=/>^'=^V"'' 


^  +  E' 
EE' 


(8). 


Eqs.  (6),  (7),  and  (8)  give  ultimate  resistances  if  w  is  the 
ultimate  intensity  of  resistance  for  the  roller. 

It  is  to  be  observed  that  the  main  assumptions  on  which 
the  investigation  is  based  lead  to  an  error  on  the  sid*'  cJ 
safety. 

If  for  wrought  iron,  w  =  12,000  pounds  per  square  inch, 
and  E  =  E'  =  28,000,000  pounds,  Eq.  (5)  gives  : 


APPENDIX    III. 

THE    SCHWEDLER  TRUSS. 

The  general  principle  applied  in  Chapter  III.  to  bowstring 
trusses,  enables  the  characteristics  of  the  Schvvedler  truss  to 
be  very  simply  shown.  In  fact,  that  truss  is  a  special  bow- 
string, having  the  least  possible  number  of  diagonal  braces 
under  the  conditions  assumed. 

Fig.  I  represents  the  elevation  of  such  a  truss,  and  the 
problem  involved  is  the  determination  of  such  depths,  near 


^ 

C \ 

K 

.    f     «     • 

/ 

^ 

'^^ 

V\ 

yv 

Xl 

"TTTPr/t- 

b 

c         d 

c 

\ 

g      'I 

) 

k 

Fig.   I. 


the  ends,  that  one  diagonal  only  will  be  needed  in  each 
panel ;  it  being  premised  that  the  inclined  web  members  or 
diagonals   are  to  sustain  tension  only. 

Let  W  —   total  (upper  and  lower  chord)  panel  fixed  load. 
"    R'   =   half  the  fixed  load  (or  weight)  of  the  bridge. 
"    w    =    panel  moving  load. 
"     /     =    length  of  span. 

"    ^     =    any  vertical  brace  or  truss  depth,  as  Cc. 
'*    dx    ~    vertical  brace  or  truss  depth,  as  Dd,  adjacent  to 

d  and  toward  centre. 
"    /      =   panel  length. 

"   «      =    inclination  of  any  diagonal,  as  Cd,  to  the  hori- 
zontal lower  chord,  i.  c,  Cdc  =  a. 

447 


448  APPENDIX  rii. 

Let  X  =  distance  from  A   to  the    intersection  of  the  pro 
longation   of  any  upper  chord  panel,  in  the 
left  half  of  the  truss,  with  the  prolongation 
to  the  left  of  the  lower  chord. 
"    J  =  the   normal    distance    from    that    point    of    inter- 
section to  the   prolongation    of  the   diagonal 
immediately    under    the    upper    chord    panel 
prolonged. 
Let  the  moving  load  pass  on  the  bridge  from  A  toward  L, 
and  let  ;/  be  the  number  of  panel-moving  loads  from  A. 

The  reaction  at  A,  for  any  position  of  the  moving  load 
will  be  : 

R  =  R^+nzv(^i-^'+J^^y    .    .     .     (I). 

Then  let  the  truss  be  imagined  divided  through  the  panel 
immediately  in  front  of  the  train. 

If  moments  be  taken  about  the  point  of  intersection  de- 
noted by  Xy  and  if  T  represents  the  tension  in  the  diagonal 
just  in  front  of  the  train,  whose  lever  arm  is  y,  there  will 
result : 

Ty  =  Rx  -  n{lV  +  w)fx +  ^-^-^^^\     .      (2). 

Eq.  (2)  is  so  written,  it  is  important  to  notice,  that  if  the 
second  member  is  greater  than  zero,  or  positive,  T  will  be 
tension.     Hence,  if  T  is  tension : 

But,    j/  =  (x  +  (n  +  l)p)  sin  a (3). 

Also,  from  similar  triangles  ; 
^-^^=—^-^.'.x  +  {n  +  i)p  =  -^^..     (4). 

By  the  aid  of  Eq.  (3) : 


THE   SCHWEDLEK    TRUSS.  ^g 

{n   +    I)/ 

—  R{7i  +  i)p  +  n{lV  +  w)       2  R  —  n{W  +w) 

~  {x  -f  (;^  +  I  )p)  sin  a  sin  a  —  * 

Hence,  by  the  aid  of  Eq.  (4)  : 

n{  W  +  wY 


(^      n{W  ^ivf. 

[R- ^ 'Yn  +  i)p 


/< 


Using  the  last  two  members  of  this  inequality  : 
d  R  —  n{PV  +  w) 


(6). 


^        ,^     ,  /                  R-  n{lV  +  zv)        \ 
Or;d>  dJi  - j^ ^^ — '- ) 


The  second  member  of  (6)  is  the  least  value  of  the  depth 
d  which  can  exist  without  inducing  compression  in  the 
diagonal  under  consideration.  This  diagonal  is  the  one 
immediately  in  front  of  the  train,  and  the  principles  given  in 
Chapter  III.  show  that  if  this  position  does  not  induce  com- 
pression, no  other  will. 

Inequality  (6)  shows  that  if  R  is  greater  than  n{lV  +  w), 
or  R>n(JV  +  zc),  d  will  always  be  less  than  d^.  U  R  = 
n{W  V  lu),  then  d=  d^. 

Again,  \i  R<i  n{  W  -!-  w),  then  will  d  be  greater  than  d^ ,  if 
tension  is  to  be  found  in  the  diagonal.  But  it  is  not  admissi- 
ble to  make  d  '>  d-^\  hence,  zvhen  n  becomes  so  great  that  R  is 
less  than  ii{  IV  +  zv),  or  R  <  n{  IV  +  zv),  T  zvill  be  compression, 
and  the  diagonal  imist  be  coiinterbraced  or  else  intersecting 
diagonals  mnst  be  placed  in  the  panels,  as  shozvn  in  Fig.  i,  near 
the  middle  of  the  truss. 

The  value  of  n  given  by 

R  <n{W  ^  zv) (7). 

2g 


.-P,  APPENDIX  III. 

will  show  the  position  of  the  head  of  the  train  when  all 
panels  between  it  and  the  centre  must  contain  intersecting 
diagonals.  All  the  other  panels  will  need  but  one  each, 
sloping  upward  and  toward  the  end  of  the  truss,  as  shown 
in  Fig.  I. 

Since  this  method  is  independent  of  the  direction  of  ap- 
proach of  the  train,  it  is  only  necessary  to  consider  one-half 
of  the  truss. 

It  is  seen  in  (6),  that  d  is  given  in  terms  of  d^,  hence  the 
latter  must  be  known  in  order  to  find  d. 

The  centre  depth  is  arbitrary  and  may  be  assigned  at  will. 
The  depths  between  the  centre  and  that  point  indicated  by 
«,  in  inequality  (7),  may  also  be  assigned  at  will ;  consequently 
^1,  next  to  the  first  "  c/"  to  be  computed,  will  be  known.  The 
first  "(^"  computed  will  be  the  "  ^/j"  for  the  next  "<'/,"  etc., 
etc.,  to  the  end  of  the  truss. 

As  a  margin  of  safety  it  will  be  well  to  make  d  a  little 
greater  than  given  by  the  second  member  of  (6). 

In  long  spans  it  would  be  well  to  make  the  truss  depth 
constant  for  a  number  of  panels  near  the  centre,  perhaps, 
even  between  the  points  given  by  (7).  This  would  make  a  con- 
siderable number  of  diagonals  and  panels  uniform  in  length, 
which  would  otherwise  lack  uniformity.  Thus  the  construc- 
tion would  be  simplified  and  cheapened. 

The  loads  have  been  taken  uniformly,  but  precisely  the 
same  methods  would  hold  if  they  were  not  uniform. 


Example. 

Let  the  following  example  (the  truss  shown    in   Fig.  i)  be 
taken  : 

Span  —  I  ■=.  (^p  —  108  feet;     .'.     /  =  12  feet. 
Centre  depth  =   16  feet. 

W  —  8.00  tons.  zv  =  1 8.00  tons. 

( W  +  zu)  =  26.00  tons. 
i?i  =  4[r  =  32.00     " 


THE    SCHWEDLER    TRUSS.  45 f 

From  Eq.  (l) : 

^=32  +  r8(«-^^)     .    .     .     (8). 

\in  =  3,  by  (8)  and  (7): 

/?  =  32  +  18  (3  -  I)  =  74  <  «(W^  +  zv)  =  78. 

Hence  the  diagonals  Dr  and  £d,  in  the  panel  in  front  of 
the  head  of  the  train,  at  ir/,  must  both  be  introduced. 

If  n  =  2,  by  (8)  and  (7)  : 

R  —  62  >  ii{  IV  +  w)  —  52. 

Hence  Cd  is  the  only  diagonal  needed  in  the  panel  CDdc, 
and  d  =  CV  is  to  be  computed  from  Eq.  (6). 

The  centre  depth  —  Ee  —  Ff  was  taken  at  16  feet  ;  let 
DdhQ  taken  at   15.5  feet. 

Since  «  =  2  ;    R  —  ti  {W  ^  w)  =  \o,  and  i?  -   ^^ — '  =  36. 

Substituting  these  values,  and  di=  15.5,  in  Eq.  (6)  : 

d  =  o.gi   dj  =  14. 1 1    feet.      Hence  let 

d—  14.5  feet  =  Cc  (Fig.  i). 

Next,  let  the  head  of  the  train  be  at  d,  i.e.,  let  «  =  i. 
Then  by  Eq.  (8) : 

y?=  32  +  16  =48. 

fi{  M/  4-  iv^ 
Also;       R  -  n{lV  +  w)  ^  22,    and    7?  - -^^ — ^I__' =  35. 

For  this  position  of  load,  d^  =  14.5  feet.      Hence,  by  Eq.  (6): 

^=  14.5(1 )  =  9-94  feet.     Hence  let 

d  —  10.00  feet  =  Bd  (Fig.  i). 


452  APPENDIX  III. 

Fig.  I  represents  the  truss,  drawn  to  scale,  with  the  various 
depths  given  or  computed  as  above,  i.  e., 

Ee  —  Ff  —  16.0  feet. 

Cc  =  Hh  —  14.5      " 
Bb  =  Kk^\o.o     " 

In  the  three  panels  adjacent  to  each  end  of  the  truss  only 
two  main  diagonals  are  thus  seen  to  be  necessary,  and  ir^ 
those  no  compression  will  ever  exist.  In  each  of  the  three 
middle  panels,  however,  two  intersecting  diagonals  will  be 
necessary,  since  no  diagonal  must  sustain  compression. 

As  is  evident,  the  expression  R  —  n{  \V  +  w)  is  the  vertical 
shear  at  the  head  of  the  train.  Hence  the  limiting  case  of 
the  inequality  (7) : 

R  =  n{W  +w), 

gives  the  two  points,  in  the  two  halves  of  the  truss,  at  which 
the  vertical  shear  at  the  head  of  the  train  is  zero.  Between 
these  points  intersecting  diagonals  or  counterbraces  are 
needed,  and  only  between  them 

After  the  truss  depths  are  fixed  by  the  preceding  method, 
the  stresses  in  the  individual  members  are  to  be  found  in  the 
usual  manner — as  in  any  other  bowstring  truss — as  shown  in_ 
Chapter  III. 


APPENDIX    IV. 

REACTIONS  AND    MOMENTS    FOR   CONTINUOUS   BEAMS. 

Case  I. — Tluee  Spans  with  Two  Intermediate  Points  of  Support  and 
Two  End  Supports. 

The  notation  of  Art.  35  will  be  used  and  reference  will  be 
made  to  Fig.  i  of  that  Art. 

M,  =  [a%^:  -  J)|  -  2/3' (A  +  4)  ip(.  -  g^  ^ 
!4  (4 +  4)  (/,  +  /,) -4"! (i). 

M,  =  -  I  J/34  +  A'2J'(^'  -  ^)f  I  -  2  (4  +  4)-    •    •  (2)- 

^■=M'-7>f (-'>■ 

Ji.  =  2P'^-^-^^-^' (4). 

^3  = 7 +  2iJ^j-~^ (5). 

In  ordinary  swing-bridges  where  /j  =  4  =  /  and   a  single 
weight  P  rests  on  /i : 

-.=H(.-;)-(--";);\-#i^,4--<^)- 


454  APPENDIX  IV. 

These  formulae  are  in  no  wise  changed  for  a  single  weight 
P  resting  on  4,  except  that  R^  and  R^  are  interchanged.  Also  : 

^^^  w -^.) /-/>(/-.)_  ^^ ^^^^ 

,'.R.^  =  P-R^-  R,^  R^ (lo). 

Case  II. — Two  Spans  with  One  Intermediate  Support. 
Reference  will  be  made  to  the  notation  and  Fig.  2  of  Art.  35. 

^■  =  ^^^0-7.)-f -■    •    •     •    C^)- 

^,  =  24-^  .i/'^.-f (,3). 

lf/,  =  /,  =  /.■ 


'2  —  M 


These  formulae  are  based  on  the  supposition  that  there 
may  be  negative  reactions  at  A  and  C  o{  Fig.  2,  Art.  35.  If 
no  negative  reactions  are  possible,  and  if  the  load  is  on  one 
arm  only,  that  arm  will  be  a  non-continuous  beam  for  such 
load,  and  the  reactions  will  be  found  by  the  simple  principle 
of  the  lever. 


APPENDIX  V. 

CANTILEVERS. 

Art.  L— Cantilever    Structures.— Positions   of  Loading  for   Greatest 
Stresses   in   the   Cantilever   Arm. 

Cantilever  structures  are  formed  of  continuous  or  semi- 
continuous  trusses  in  which  the  reactions  and  stresses  are 
equally  determinate  with  those  in  non-continuous  trusses. 
Fig.  I,  PI.  XIII.,  typifies  a  cantilever  structure  with  an 
anchor  arm  at  each  end.  The  anchorage  at  A  holds  the 
anchor  arm  ;/  in  position  under  all  conditions  of  loading,  thus 
enabling  the  cantilever  arm  ;//  to  sustain  the  suspended  span 
/.  The  latter  is  a  simple,  non-continuous  truss  with  inclined 
end-posts,  suspended  at  the  lower  extremities  of  the  vertical 
tension  members  at  C  and  D,  thus  readily  allowing  expansion 
and  contraction  to  take  place  both  in  the  suspended  span 
itself  and  in  the  cantilever  arms  adjacent  to  it.  As  can- 
tilevers are  erected  without  lower  false-works  between  the 
piers  Band  E,  each  cantilever  arm,  with  the  adjacent  half  of 
the  suspended  span,  is  built  out  from  each  of  those  main 
piers  ;  hence  the  members  FG  sustain  compression  during 
erection,  while  HD  take  tension.  These  erection  stresses 
are  frequently  very  heavy,  and  the  members  themselves  must 
be  designed  to  take  them.  At  the  same  time,  proper  details 
must  be  arranged  so  that  after  the  structure  is  complete  the 
requisite  expansion  and  contraction  can  take  place  at  each 
end  of  the  suspended  span.  These  results  are  usually  accom- 
plished by  folding  wedges,  or  wedges  and  rollers,  removed 
after  erection,  in  connection  with  oblong  pin  holes. 

Similarly,  nearly  the  whole  of  the  upper  chord  of  the  sus- 
pended span  must  be  designed  to  take  the  tensile  erection 
stresses,  and  nearly  the  whole  of  the  lower  chord  the  corre- 
sponding compression  erection  stresses.  The  erection  stresses 
in   the   cantilever  arms  arc,  of  course,  the   same   in  kind  as 


45^  APPENDIX    V.  * 

those  caused  by  the  moving  and  fixed  loads,  and  no  reversion 
takes  place. 

The  conditions  of  loading  for  the  greatest  stresses  in  the 
suspended  span,  in  addition  to  the  erection  stresses,  are  pre- 
cisely the  same  as  those  for  the  ordinary  non-continuous 
truss  given  in  detail  in  Art.  7,  and  they  will  receive  no  fur- 
ther attention  here. 

Greatest    Web  Stresses  in   Cantilever  Arm. 

In  seeking  the  greatest  web  stresses  in  the  cantilever  arm, 
reference  will  be  made  to  Fig.  i,  PI.  XIII.,  and  a  perfectly 
general  system  of  loading  will  be  assumed.  The  chords 
of  the  cantilever  arm  will  also  be  assumed  to  be  not  parallel. 
Let  the  greatest  stress  5,  in  MN,  be  desired,  and  let  the 
system  of  weights  or  concentrated  loads  W{,  W2,  etc.,  extend 
over  the  entire  suspended  span  /,  and  over  the  cantilever  arm 
;«,  from  C  to  some  point  to  the  left  of  M.  Then  let  i  rep- 
resent the  distance  from  C  to  the  point  of  intersection  of  the 
chord  sections  in  the  same  panel  with  MN,  while  A  is  the 
length  of  the  normal  dropped  from  the  intersection  point  to 
MjV  prolonged.  Also  let  ^1  represent  the  distance  from 
D  to  the  centre  of  gravity  of  the  load  on  /;  ^9  the  distance 
from  the  same  point  to  the  centre  of  gravity  of  the  load  to 
the  left  of  C;  g^  the  similar  distance  from  M  ior  the  load 
to  the  left  of  M)  and  ^,  the  similar  distance  from  M  of  the 
loads  Wn,  W^,  etc.,  in  the  panel  in  question.  The  loads  on 
the  left  of  M  are  W^,  W2,  etc.  The  distance  from  C  to  M 
is  represented  by  k.  R  is  the  downward  reaction  or  upward 
pull  on  the  anchorage  at  the  extremity  of  n ;  while  R^  is  the 
upward  reaction  at  the  main  pier.  The  weights  over  the 
various  portions  of  the  structure  will  be  represented  as 
follows : 

2  W  for  total  weights  in  /. 


CANTILEVERS,  457 

The  centre  of  gravity  distances,  gy,  g.,,  and  g^,  can  be  read- 
ily found  by  aid  of  tabulations  like  those  given  in  Arts.  9  arid 
II,  while  ^4  will  be  found  by  taking  moments  of  IV^,  W^,  etc., 
about  M. 

Then,  by  moments  in  the  ordinary  manner: 

B  ^'^klV^{^^i±±^^^lV (I). 

R.,  =  reaction  at  C  from  load  on  suspended  span  =^  2W    (2). 

y?,^y?+i?,+5j^=^(^^i)in^^4-(^^'±^-i)SjF(3). 

If  the  length  of  the  panel  in  question  is/,  then  that  portion 
of  the  loads  l\\,  JV^,  etc.,  resting  in  it  and  carried  to  M,  is: 

(w,  ^  rr,  +  etc.)  (i  -  ^) (4). 

If  the  cantilever  arm  be  now  supposed  divided  through 
the  panel  MN,  and  if  the  moments  about  intersection  of 
chords  be  taken  of  all  the  forces  external  to  that  portion  of 
the  structure  to  the  left  of  this  line  of  division,  including 
the  reactions  R  and  —  R^,  there  will  result  : 

5  =  -,  I  ^'  i^r  +  (^,  -  /  4-  /)  1 JF 

-  (f-Fi  +  W.^-  etc.  +  W,  +  fr;  +  etc.)  {k  +  i) 

-{W,^  W,  +  etc.)^-3  +  (^^4  +  ^^'5  +  etc.)(/J'  +  ifj  [  .     .  (5). 

In  order  that  5  may  be  a  maximum  or  a  minimum,  AS 
must  be  zero  when  gx,  gi,  gs,  and  gi  each  varj'  by  the  same 
small  amount,  as  when  the  loads  are  all  advanced  slightly  to 
the  left.  It  is  to  be  remembered,  however,  that  the  variation 
of  ^4  will  be  opposite  in  sign  to  that  of  the  others.     By   thus 


458  APPENDIX    V. 

making  AS  equal  to  zero,  there  will  result  as  the  desired  con- 
dition  : 

W,+  IV,  +  etc.  +  {W,  +  W,  +  etc.)  ^-^^  =  \2W  +  S:W. 

P        .    ^ 

.-.  ( W,  +  Wr,  +  etc.)  ^^-^^^  =  -^ i IV  +  i IV.     .     (6). 

In  case  there  may  be  a  number  of  maxima  indicated  by 
Eq.  (6),  the  greatest  must  be  sought  by  trial.  Eq.  (6)  shows, 
as  might  have  been  anticipated,  that  the  loads  IV^+  ff^a  +  etc, 
between  the  panel  in  question  and  the  main  pier,  do  not 
directly  affect  the  conditions  for  greatest  stress. 

If  the  load  is  of  the  uniform  intensity  w,-d,Vidi  x  is  the  por- 
tion of  the  panel/  covered  by  it,  Eq.  (6i  is  only  satisfied  by 
making  x  =  /.  This  result  might  also  have  easily  been 
anticipated,  for  it  is  clear  that  if  the  load  is  uniform  it  must 
at  least  reach  to  M,  for  the  panel  MN.  The  amount  by 
which  it  may  extend  to  the  left  of  M  is  a  matter  of  indiffer- 
ence. 

If  the  chords  are  parallel,  i  is  infinitely  great  and  Eq.  (6) 
becomes  : 

^4+    IV, -^  etc.  =  j  2 IV.  .     .     .     .     (7). 

In  the  general  case,  when  Eq.  (6)  has  fixed  the  proper 
position  of  loading,  the  corresponding  values  of  ^i,  £:,,  g^,  and 
g^  can  be  at  once  obtained  by  the  methods  already  indicated, 
and  Eq.  (5)  will  then  give  the  desired  greatest  value  of  5.  In 
the  case  of  a  uniform  load,  u>,  that  equation  takes  a  much 
simplified  form,  since : 

W^  +  W2+  etc.  =  o ;  Wi  +  IV,  +  etc.  =  wp ; 


I 


2W=wk;  2JV=wl; 


^  ,    .     ^  A  P 


CANTILEVERS.  459 

Substituting  these  values  in  Eq.  (5) : 

If  the  chords  are  parallel  and  a  is  the  angle  between  ^'  and 
a  vertical  line,//  =    (00  =  z)  cos  ix\  hence: 

S  =  zv  i~  -\-  k  —^  \  sec  IX (9). 


With  the  same  uniform  load,  w,  per  lineal  foot,  and  with 
kw  on  the  cantilever  arm  in,  Eqs.  (i),  (2),  and  (3)  become : 

R    =^"l(Uk-^\ (I0\ 

n  \2  2mJ 

/?,  =  '^' (.1). 

2 

„        wl  fm         \  ,  fm        k  \  ,     . 

R^  =  —  [^+\]+wk[ —  +  I  ) (12). 

2  \n         J  \n       2n         J  ^     ' 

If  the  load,  w,  covers  the  entire  arm  ;;/,  k  =  ;;/,  and  : 

R^w-{l+vi) .     .     .     (13). 

2n  '  ^ 

wl        ,   „        %vl  (m   ^      \  f  m   ^       \        I     . 

R,  =  --,  and  ye,  =  ^  (-  +  l)  +  wm  (^-  +  .  ).      (.4). 

If  a  single  weight,  W,  rests  on  the  suspended  span,  /,  at  the 
distance  ^,  from  D  : 

^=,K'-M.;^,=  ^,and^,  =  i^('^.).    .     (.5). 

If  a  single  weight,  W,  rests  on  the  arm  m  at  the  distance  k 
from  C: 

ji^KiaiJzJ);  ji.  =  o:  and  R,  =  w('^i^+  i).     (16). 


.460  APPENDIX    V. 

Greatest  Chord  Stresses  in  the  Cantilever  Arm. 

Reference  will  again  be  made  to  Fig.  i  of  PL  XIII.,  in 
which  the  compression  web  members  are  vertical ;  the  nota- 
tion will  remain  as  before.  The  greatest  chord  stress,  S^.,  in 
any  panel,  as  MN,  will  be  sought.  Let  //'  represent  the 
normal  dropped  from  M  on  the  chord  panel  whose  greatest 
stress  is  S^. ;  then,  if  moments  be  taken  about  M,  h^  will  be 
the  lever  arm  of  Sc,  and  there  will  result : 

5c  =    .-J  I  Rx  {m  —  k)  —  R  {in  +  n  —  k) 

-(W,  +  JV,  +  ctc.)s,^.    .     .     .     (17)- 

.-.  5,  =  i  U^^W-  ^j^2W-  {/+  k)2lV  . 

-{IV,+  W^, +  etc.)^,|  ....     (18). 

For  a  maximum  or  minimum  ^Sc  =  o  when  gx,  g^,  and  g^  are 
the  only  variables  in  the  second  member  of  the  equation  ; 
hence : 

-'^w^  ^w-{w,  +  w,  +  t\.c.)  =  hiv.  .   .   (19). 

This  condition  for  a  maximum  is  seen  to  be  independent  of 
i;  hence  it  is  precisely  the  same  whether  the  chords  are 
parallel  or  inclined.  After  the  values  of  g,  gg,  and  ^3,  corre- 
sponding to  the  position  of  loading  fixed  by  Eq.  (19),  are  in- 
serted in  Eq.  (18),  the  desired  chord  stress  will  at  once  result. 
It  is  to  be  borne  in  mind,  however,  that  there  may  be  several 
maxima  fixed  by  Eq.  (19),  and  that  the  greatest  of  these,  to 
be  found  by  trial,  is  the  "  greatest  stress  "  desired. 

If  the  load  is  of  the  uniform  amount,  zi>,  per  lineal  unit  of 
structure,  Eq.  (19)  is  satisfied  only  by  making  ^'rF=  kw; 
hence,  the  moving  load  iniist  cover  the  suspended  span,  and  the 
cantilever  arm  from  its  end  to  the  vertical  line  through  the  cen- 


CANTILEVERS. 


461 


tre  of  Diomcnts — i.e.,  the  distance  k.     Eq.  (18)  then  takes  the 
value,  for  uniform  loading  : 


5. 


wk 

2h' 


{k  +  I). 


(20). 


For  a  panel  in  the  horizontal  chord,  //'  is  simply  the  depth 
of  truss  at  the  origin  of  moments. 

In  case  all  web  tnembers  arc  inclined,  as  indicated  in  Fig.  i, 
so  that  any  lower  chord  panel  point  M^  is  at  the  horizontal 
distance  q  from  that  in  the  upper  chord,  Eq.  (18)  takes  the 
form  : 

5e  =  I  \g!^W-  (i±^ii'fr-(7+  k  ^-  q)'kw 

-(W^,  +  ^F, +  etc.)(/-^)"^|  .     .     .     .     (21). 


Again  making  ^5^=  o,  with  g^,  g.,,  and  g.^  the  only  vari- 
ables, there  results  for  a  maximum  or  minimum  : 


f  I^  =  i±^il^+  {W,  +  W,  +  etc.)^^ ^ 


Or, 


/ 


^^2W=^^W-{IV,+   lV,  +  etc./-^-^, 


(22). 


(23). 


If  ^  =  o,  Eq.  (19)  at  once  results.  The  observations  fol- 
lowing Eq.  (19),  regarding  /,  gi,g2,  and  gs,  obviously  hold  true 
for  this  case  also.     If  the   moving  load  is  of  the  uniform  in- 


462  APPENDIX    V. 

tensity  w,  Eq.  (23)  is  satisfied  only  by  making  it  cover  k  -\-  p\ 
hence  that  equation  takes  the  form  : 

k  +  q=^k+^^  =  k+q;    .      .     .     .     (24). 

which  shows  that  t/ie  laiiform  moving  load  imist  cover  the  sus- 
pended span,  and  the  cantilever  arm  from  its  end  to  the  further 
extremity  of  the  panel  in  which  the  chord  stress  is  sought. 
For  this  uniform  moving  load  Eq.  (21)  takes  the  form: 


-^k^q){k-^p  +  -y^-^p-q)^-^y     (25)., 


h'  \       2 


If  ^  z=  o  in  this  equation,  Eq.  (20)  immediately  follows. 

If  the  panel  in  the  lower  chord  is  under  consideration,  M, 
Fig.  I,  becomes  the  origin  of  moment,  and  as  the  rhoving 
load  is  supposed  to  traverse  the  upper  chord,  the  conditions. 
for  a  maximum  and  the  corresponding  formulae  are  precisely 
the  same  as  for  the  case  with  vertical  compression  web  mem- 
bers. 

If,  with  all  inclined  web  members,  the  moving  load  tra- 
verses the  lower  chord,  the  preceding  conditions  and  formulae 
apply  precisely  as  they  stand.  It  is  only  to  be  borne  in 
mind  that  the  formulae  involving  q  are  to  be  used  for  the 
chord  carrying  the  moving  load. 

If  the  tension  members  are  vertical,  and  the  compression 
members  inclined,  as  in  the  Howe  truss,  or  as  exemplified  by 
the  inclined  post,  P,  Fig.  i,  of  PI.  XIII.,  the  positions  of 
moving  load  for  greatest  chord  stresses  will  be  at  once  found 
by  making  q  —  p  in  Eq.  (23),  which  gives: 

—j^^W=2W (26). 

for  the  general  load,  and  the  same  condition  for  a  uniform 
load  as  that  described  in  italics  immediately  after  Eq.  (24). 
By  making  q  —  p,  in  Eq.  (25),  the  expression  for  the  greatest 
chord  stress  under  a  uniform  load,  w,  becomes — 


CA  A'  riLE  VERS.  4^3 

"  //'    (  2  2  )  2ll'-  '         ' 

If  the  truss  is  of  uniform  depth,  /^'  is  that  depth. 

The  preceding  investigations  hold  true  for  any  system  of 
triangulation  or  for  any  system  of  loading.  They  do  not 
apply  to  multiple  systems  of  bracing,  as  such  systems  do  not 
admit  of  exact  calculation  of  stresses.  When  they  are  used, 
each  system  must  be  assumed  to  act  independently  of  all  the 
others  (although  the  assumption  cannot  be  definitely  estab- 
lished) and  to  Z2sry  the  system  of  maxima  concentrations 
permitted  by  the  moving  load  used,  or  else  such  a  uniform 
load  as  will  be  practically  equivalent  to  the  real  load. 

Although  there  is  not  at  the  present  time  (1890)  any  con- 
structive or  other  proper  reason  for  the  use  of  multiple  sys- 
tems of  bracing  in  any  pin  structure  of  proportions  hitherto 
employed,  the  above  assumptions  are  \\ithin  the  limits  of 
safety,  and  may  be  used  where  necessary  and  unavoidable. 

Another  constructive  device  is  that  shown  in  Fig.  2.  It 
may  be  used  advantageously  in  cases  where  the  cantilevers 
are  supported  on  high  iron  or  steel  piers  or  towers,  when  the 
latter  would  become  too  expensive  if  constructed  of  masonry 
of  corresponding  dimensions.  It  leads  to  appreciable  econ- 
omy by  the  shortening  of  the  clear  cantilever  opening,  as 
well  as  the  anchor  arm.  It  consists  in  separating  the  feet  of 
the  two  inclined  posts,  P  and  IB' ,  of  Fig.  r,  PI.  XIII.,  by  some 
convenient  distance,  o,  and  omitting  all  bracing  in  the  rect- 
angle, BB'B^B^,  Fig.  2,  so  that  no  shear  can  be  transferred 
past  B'  to  the  left,  or  past  B^  to  the  right.  Since  this  last 
condition  exists,  the  reaction,  R,  and  the  positions  of  loading 
for  all  the  maxima  web  and  chord  stresses  in  the  cantilever 
arm,  as  given  by  the  preceding  formulae,  will  hold  for  this 
case  without  any  change  whatever.  It  is  only  to  be  carefully 
observed  that  ///  and  ;/  are  to  be  measured  from  the  extremi- 
ties of  the  cantilever  and  anchor  arms,  respectively,  to  the 
extremities  of  the  open  panel,  0,  as  shown  in  Fig.  2.  The 
reaction,  R^  at  B\  will  equal  the  sum  of  the  vertical  compo- 
nents of  all  the  inclined  stresses  in  the  first  panel  of  m — i.e., 


464 


APPENDIX    V. 


the  total  shear  in  tliat  panel,  added  to  the  panel  loads  acting 
at  B  and  B .  Similarly,  the  reaction  R"  at  B^  will  equal  the 
sum  of  the  vertical  components  of  all  the  inclined  stresses 
in  the  first  panel  of  n — i.e.,  the  total  shear  in  that  panel, 
added  to  the  panel  loads  acting  at  B^  and  B^.  Finally,  the 
reactions,  7?,'  and  7?/',  of  Fig.  2,  added  together,  will  equal 


Fig.  3. 


the  reaction,  7?„  of  Fig.  i,  PI.  XIII.  Thus  it  is  seen  that 
the  formulae  already  established  for  the  latter  case  will  meet 
the  former  without  any  change  whatever. 

The  case  of  the  subdivided  panel  shown  in  Fig.  3  is  also 
covered  by  the  preceding  formulae.  The  general  and  detailed 
considerations  governing  their  applications  are  precisely  the 
same  as  those  given  at  length  in  Art.  7  ;  it  is  not  necessary, 
therefore,  to  give  them  further  attention  here,  except  to  ob- 
serve that  for  the  greatest  stress  in  be,  the  panel  length  to  be 
taken  is  ad,  while  ae  is  the  panel  for  the  greatest  stress  in  ab. 

Art.  2. — Positions  of  Moving  Load  for  Greatest  Stresses  in  the  Anchor 
Arm,  and  the  Greatest  Stresses  Themselves. 

No  load  placed  on  the  anchor  arm  of  a  cantilever  structure 
will  affect  in  any  way  the  reaction  or  any  shear  in  the  canti- 
lever arm,  except  such  as  may  result  from  its  slight  deflectional 
movement,  being  too  small  to  be  expressed  in  appreciable 
amounts.  It  at  once  follows  from  this  fact  that  all  stresses 
in  the  anchor  arm,  due  to  loads  on  that  arm,  may  be  compu- 
ted precisely  as  if  it  {AIB'L  of  Fig.  i,  PI.  XIII.)  were  a  simple, 
non-continuous  truss  supported   at  each    end  {i.e.,  A  and  B' 


CANTILEVERS.  465 

of  Fig.  I,  PI.  XIII.).  The  greatest  of  these  stresses  so  found 
are  then  to  be  combined  with  the  greatest  stresses  of  the 
same  kind  due  to  the  load  on  the  cantilever  arm  and  result- 
ing from  the  reaction,  R,  at  the  anchorage  ;  this  combination 
will  give  the  resultant  greatest  stresses  desired. 

Although  the  preceding  general  observations  are  compre- 
hensive and  sufficient  for  the  determination  of  all  the  greatest 
stresses  desired,  it  will  be  well  to  extend  them  in  some  detail, 
without,  however,  establishing  any  formulae. 

Inasmuch  as  all  upper  chord  stresses  in  the  anchor  arm,  due 
to  load  on  that  arm  only,  will  be  compressive,  it  is  clear  that 
the  greatest  tension  in  the  upper  chord,  so  far  as  it  may 
exist,  will  be  found  with  no  moving  load  on  the  anchor  arm, 
and  with  the  moving  load  so  placed  on  the  cantilever  arm  as 
to  give  the  maximum  bending  moment  (and,  hence,  chord 
stresses)  over  the  main  pier  between  the  anchor  and  cantilever 
arms.  This  position  of  the  moving  load  on  the  cantilever 
arm  can  be  found  for  the  various  cases  from  Eqs.  (18),  (20\ 
(21 ),  and  (25)  of  the  preceding  Art.  The  resulting  moment 
over  the  main  pier,  divided  by  n,  will  give  the  reaction,  R,  with 
which  the  greatest  tension  throughout  the  upper  chord  and 
the  greatest  compression  throughout  the  lower  chord  are  to 
be  found — i.e.,  so  far  as  this  tension  or  compression  exists. 

If  the  anchor  arm  is  very  long  in  comparison  with  the  can- 
tilever arm,  the  fixed  load  compression  in  the  upper  chord 
and  tension  in  the  lower  may  be  so  great  that  there  will  be 
but  little  upper  chord  tension  and  lower  chord  compression, 
even  for  the  conditions  producing  maxima. 

Inasmuch  as  all  loads  on  the  suspended  span  and  cantilever 
arms  produce  tension  in  the  upper  chords  of  the  anchor  arms 
and  compression  in  the  lower,  the  positions  of  moving  load 
for  the  greatest  compressions  in  the  upper  chord  of  either 
anchor  arm,  or  greatest  tensions  in  the  lower,  are  precisely 
the  same  as  if  it  (the  anchor  arm)  were  a  simple,  non-continu- 
ous truss,  the  cantilever  arms  and  suspended  span,  at  the 
same  time,  carrying  no  moving  load.  The  greatest  compres- 
sive stresses  in  the  upper  chord  and  tensile  stresses  in  the 
lower  chord,  thus  found  in  the  anchor  arm,  as  for  a  simple, 


466  APPENDIX    V. 

non-continuous  span,  are  to  be  combined  with  the  upper 
chord  tension  and  lower  chord  compression  produced  by  the 
fixed  load  of  the  cantilever  arms  and  suspended  span,  whence 
will  result  the  maxima  chord  stresses  desired. 

The  greatest  shears  in  the  anchor  arm  will  evidently  be 
affected  by  the  magnitude  of  the  reaction  /?,  Fig.  i,  PI.  XIII., 
at  its  extremity.  If  a  downward  shear,  producing  tension  in 
members  sloping  similarly  to  OQ,  or  compression  in  IB\  is 
called  a  main  shear,  then  any  condition  of  loading  on  the 
structure  which  will  increase  the  downward  reaction,  R,  will 
increase  the  main  shears  and,  hence,  the  stresses  in  the  main 
web  members,  as  AL,  LV,  UT,  etc.,  and  IB'  will  be  called. 
But  the  maximum  value  of  the  downward  moving  load  reac- 
tion or  pull  at  the  extremity  of  the  anchor  arm  will  be 
found  with  no  moving  load  on  the  latter,  and  for  the  maxi- 
mum value  of  the  bending  moment  over  the  main  pier  B',  as 
given  by  Eqs.  (i8),  (20),  (21),  and  (25),  of  the  preceding  Art., 
as  already  stated  on  page  465.  Finally,  that  part  of  the 
maximum  main  shear  in  any  panel  of  the  anchor  arm  due  to 
the  moving  load  on  that  portion  of  the  structure  alone,  is 
found  precisely  as  if  it  were  a  simple,  non-continuous  truss 
with  the  given  system  of  loading,  advancing  from  the  extrem- 
ity of  the  anchor  arm  to  the  main  pier.  The  positions  of 
moving  load,  therefore,  for  the  main  shears  or  main  web 
stresses  are,  first,  so  much  on  the  cantilever  arms  and  sus- 
pended span  as  will  produce  the  maximum  downward  pull  at 
the  extremity  of  the  anchor  arm,  and,  second,  with  this  con- 
stant condition  in  the  main  span  of  the  structure,  an  advanc- 
ing moving  load  from  the  extremity  of  the  anchor  arm, 
treated  as  a  simple,  non-continuous  span,  to  the  main  pier. 
The  maximum  moving-load  shears  or  stresses  thus  found,  com- 
bined with  the  fixed-load  shears  or  stresses  in  the  anchor  arm, 
will  give  the  resultant  greatest  web  member  stresses  desired. 

It  may  sometimes  happen  that  the  anchor  arm  will  be  so 
long  that  the  greatest  downward  pull  at  its  extremity  due  to 
the  moving  load  will  be  less  than  an  existing  upward  reac- 
tion due  to  the  fixed  load.  In  such  a  case  no  anchorage,  of 
course,  will  be  required,  as  the  reaction  at  the  extremity  of 


CA  N  TIL  E  VERS.  4^7 

the  anchor  arm  will  always  be  upward.  The  preceding  con- 
ditions for  greatest  main  web  stresses,  however,  hold  in  all 
cases  without  exception. 

The  greatest  counter  shears  and,  hence,  stresses  in  counter 
web  members  such  as  QV,  or  VL  if  under  compression,  or 
AL  in  tension,  Fig.  i,  PI.  XIII.,  will,  for  the  same  general  rea- 
sons just  stated  in  connection  with  the  greatest  main  shears,  be 
found  under  the  moving  load  advancing  from  the  main  pier 
B'  to  the  extremity  of  the  anchor  arm,  under  the  supposition 
that  the  latter  is  a  simple,  non-continuous  truss,  with  no  mov- 
ing load  whatever  on  the  cantilever  arms  or  suspended  span. 

If  a  portion  of  the  web  members  are  vertical  posts,  as 
shown  in  Fig.  i  of  the  plate,  so  that  the  kind  of  stress  in 
them  is  the  same  whether  acting  as  counters  or  main  web 
members,  those  nearest  the  extremity  of  the  anchor  arm  will 
probably  take  their  greatest  stresses  as  counters,  but  those 
more  remote  are  doubtful  and  must  be  computed  both  as 
main  and  counter  members  in  order  to  find  the  greatest 
values.  Those  near  the  main  pier  will  take  their  greatest 
stresses  as  main  web  members. 

Art.  3. — Positions  of  Moving  Load  for  Greatest  Stresses  in  Anchor 
Spans  and  the  Greatest  Stresses  Themselves. 

The  anchor  span  is  shown  as  the  span  ;/  in  Fig.  2,  of  PI. 
XIII.,  on  the  left  of  which  there  is  supposed  to  be  a  cantilever 
arm  similar  to  vi  on  the  right.  The  anchor  span  is  always  of 
such  length  and  weight  that  the  reactions,  R  and  Ry,  are  invari- 
ably upward.  Hence  the  anchor  span  is  identical  in  character 
with  the  anchor  arm,  whose  length  and  weight  are  so  great 
that  the  reaction  at  its  extremity  over  the  anchorage  is 
always  upward.  Hence  the  positions  of  moving  load  for  the 
greatest  stresses  are  precisely  the  same  as  those  established 
for  the  anchor  arm  in  the  last  Art. 

The  greatest  moving-load  stresses  in  the  span  ;/,  Fig.  2,  PI. 
XIII.,  are  to  be  determined  precisely  as  if  the  trusses  were  of 
a  simple,  non-continuous  character  ;  and  the  fixed-load  stresses 
are  to  be  found  in  the  same  trusses  under  exactly  the  same 
conditions.     The  greatest  difference  of  moments  at  the  two 


468  APPENDIX    V. 

sections,  BR^  and  AR,  is  then  to  be  found  and  divided  by  the 
span  length  n,  thus  giving  a  reaction  ivJiich  is  to  be  treated  pre- 
ciscly  as  the  downward  reaction  at  the  extremity  of  the  anchor 
arm  in  the  preceding  article.  This  downward  reaction  will  pro- 
duce stresses  in  the  chord  and  web  members  of  the  anchor 
span,  which  are  next  to  be  determined  in  the  usual  manner  for 
a  load  hanging  at  the  extremity  of  the  span  n,  with  the  other 
extremity  fixedly  held.  If  both  cantilever  arms  flanking  the 
span  n  are  of  the  same  length  ni,  and  of  the  same  weight 
similarly  disposed,  the  fixed-load  moments  at  the  sections 
AR  and  BR^  will  always  be  equal  (but  not  otherwise),  and 
the  difTerence  in  moments  mentioned  above  will  be  entirely 
due  to  the  moving  load  on  one  of  the  adjoining  cantilever 
spans.  Any  difference  of  moments  at  those  two  sections 
resulting  from  dissimilarity  or  inequality  of  fixed  loads  must 
be  added  to  or  subtracted  from  the  unbalanced  moving-load 
moment  just  described.  Finally,  the  fixed-load  balanced  mo- 
ment at  either  of  the  sections  AR  or  BR^  is  then  to  be  deter- 
mined and  divided  by  the  depth  of  the  truss  in  order  to  find 
the  uniform  tension  throughout  the  length  of  the  upper 
chord,  and  the  uniform  compression  throughout  the  length 
of  the  lower,  if  the  truss  has  a  uniform  depth.  The  stresses 
resulting  from  these  four  sources — viz.,  the  moving  load  and 
the  fixed  load  on  the  span  n\  the  unbalanced  moment  over 
the  piers  due  to  possibly  both  moving  and  fixed  loads,  and 
the  balanced  moment  due  to  fixed  load  only — are  to  be  so 
combined  as  to  produce  maxima  in  each  and  all  the  truss 
members. 

The  greatest  web  stresses  will  require  the  greatest  unbal- 
anced moving-load  moment  alternately  at  each  end  of  the 
anchor  span,  with  the  usual  progressive  movement  of  moving 
load  for  ordinary  non-continuous  trusses ;  but  the  greatest 
upper  chord  compression  and  lower  chord  tension  will  be 
found  with  no  moving  load  on  the  adjacent  cantilever  arms 
or  suspended  spans. 

The  greatest  upper  chord  tension  and  lower  chord  com- 
pression will  be  found  by  combining  the  fixed-load  stresses, 
found    as   indicated    above,  with    the    moving-load   stresses 


CANTILE  VERS.  469 

induced  by  the  greatest  possible  moving-load  moments 
simultaneously  at  each  end  of  the  anchor  span,  the  latter,  at 
the  same  time,  being  entirely  free  from  moving  load. 

If  the  anchor  span  is  not  of  uniform  depth,  the  balanced 
fixed-load  moments  at  its  extremities  will  induce  stresses 
in  the  web  members,  which  are  to  be  determined  in  addition 
to  those  in  the  chords  for  combination  with  the  other 
stresses  described  above  in  the  search  for  the  maxima. 

Art.  4. — Wind    Stresses. 

The  stresses  due  to  the  action  of  the  wind  on  the  various 
portions  of  a  cantilever  structure  play  a  very  important  part 
in  its  design.  The  conditions  of  wind  loading  are  much  more 
varied  than  in  ordinary  non-continuous  spans,  and  the  result- 
ing stresses  must  be  computed  with  great  care  and  thorough- 
ness. The  pressures  against  the  different  members  of  the 
structure  constitute  a  fixed  load,  and  all  resulting  stresses  are 
to  be  found  by  precisely  the  same  general  methods  as  given 
in  the  preceding  articles  for  the  vertical  fixed  loads — i.e.,  own 
weights. 

Fig.  5  of  PI.  XIII.  shows  the  horizontal  system  of  lateral 
bracing  between  the  horizontal  upper  chords,  ABCDE  of 
Fig.  I,  If  the  wind  blows  on  the  entire  structure  in  a  given 
direction,  as  shown  by  the  arrow  in  Fig.  5,  there  will  in  gen- 
eral be  a  horizontal  reaction  induced  at  A,  in  direction  and 
amount  depending  on  the  relative  proportions  of  anchor  and 
cantilever  arms  and  suspended  truss.  This  reaction  must  be 
provided  for,  at  A,  by  a  proper  connection  between  the  end 
of  the  anchor  arm  and  masonry  at  that  point,  so  designed 
that  the  requisite  longitudinal  expansion  and  contraction 
may  at  the  same  time  take  place.  The  amount  of  this  reac- 
tion is  to  be  determined  precisely  as  the  fixed-weight  reac- 
tion, R,  at  the  same  point,  in  Fig.  i.  This  reaction,  R,  will 
be  composed  of  two  parts,  one  of  which  is  due  to  the  wind 
pressures  along  the  upper  chord,  ABCD,  Fig.  i,  and  the  other 
to  those  along  the  lower  chord,  ALB'NG;  and  each  is  to  be 
found  as  indicated  above.  By  means  of  these  horizontal 
reactions  at  A,  and  the  panel  wind  pressures  at  the  upper 


4/0  APPENDIX    V. 

and  lower  chord  points,  all  the  fixed-load  wind  stresses  in 
the  upper  and  lower  lateral  systems,  both  of  which  are  pro* 
jected  in  Fig.  5,  may  be  at  once  completely  established. 

The  wind  pressures  against  the  moving  train  form  a  mov- 
ing load  in  those  chords  carrying  the  train  (the  upper  chords 
in  Fig.  I,  PI.  XIII.,  and  the  intermediate  and  lower  chords  in 
Fig.  3).  The  conditions  under  which  these  moving  wind 
loads  produce  their  horizontal  reactions  at  A,  and  their  great- 
est stresses  in  the  web  and  chord  members  of  the  lateral  sys- 
tem, are  precisely  the  same  as  those  described  in  detail  for 
the  vertical  or  train  moving  loads  in  the  preceding  Arts. 
The  cantilever  conditions,  so  to  speak,  including  span  lengths 
for  the  former  loads,  are  identical  with  those  for  the  latter. 
Hence  the  positions  for  the  wind  maxima  must  be  deter- 
mined by  exactly  the  same  general  methods  as  are  used  for 
the  vertical  loads.  The  computations  for  the  moving  wind 
stresses  are  simplified  by  the  fact  that  that  load  is  of  uni- 
form intensity. 

The  combination  of  the  fixed  and  moving  wind  load 
stresses  according  to  the  usual  methods  will  give  the  desired 
resultants  in  all  the  members. 

In  the  case  of  the  anchor  span  of  Fig.  2,  PI.  XIII.,  the  same 
general  observations  as  those  already  made  are  to  be  applied. 
The  fixed  wind  loads  on  the  span  are  to  be  treated  precisely 
like  the  fixed  vertical  loads,  and  the  resulting  wind  stresses 
in  the  upper  and  lower  lateral  systems  are  to  be  computed 
accordingly.  Again,  the  moving  wind  loads  accompany  the 
moving  vertical  or  train  loads,  and  all  the  conditions  deter- 
mining the  greatest  stresses  for  the  latter  determine  those 
for  the  former  also.  The  same  set  of  greatest  moving  wind 
stresses  must  therefore  be  found  as  for  the  moving  vertical 
loads,  and  followed  by  their  combination  with  the  fixed  wind 
stresses,  in  order  to  reach  the  resultant  wind  stresses  desired. 

Inasmuch  as  it  is  known  that  the  highest  wind  pressures 
cover  comparatively  small  areas,  it  will  usually  be  necessary 
to  supplement  the  preceding  wind  computations,  which  are 
based  on  the  assumption  that  the  wind  presses  equally  over 
the  entire  structure,  by  others  resulting  from  the  application 


CA  JV  TILE  VERS.  47 1 

of  the  highest  pressure  to  some  particular  portion,  as  the 
clear  cantilever  span  or  the  anchor  span,  or,  possibly,  the 
anchor  arm,  and  provide  for  the  resulting  stresses. 

The  extent  to  which  these  supplementary  and  special 
computations  are  to  be  made  will  depend  upon  the  judgment 
of  the  engineer  acting  on  the  circumstances  of  each  case. 

The  transverse  bracing  in  the  vertical  and  inclined  planes 
of  the  various  pairs  of  web  members  is  to  be  designed  under 
the  same  assumptions  of  wind  transference,  and  in  accord- 
ance with  the  same  principles  used  for  the  same  general 
purposes  in  ordinary  non-continuous  spans.  The  wind  loads 
in  the  upper  chords  of  the  suspended  trusses  of  the  cantilever 
span  should  always  be  carried  down  to  the  lower  chords  of 
the  cantilever  arms  at  their  extremities,  and  along  those 
chords  to  the  piers,  in  order  that  the  overturning  wand  effect 
on  the  cantilever  and  anchor  arms  may  be  reduced  to  a  mini- 
mum. In  the  same  manner,  the  wind  loads  in  the  upper 
chords  of  the  cantilever  and  anchor  arms,  which  are  resisted 
by  the  main  piers,  should  be  carried  down  to  them  by  the 
transverse  bracing  between  the  inclined  posts  acting  as  por- 
tals at  those  places.  In  high  cantilever  trusses  the  overturn- 
ing wind  effect  is  frequently  a  very  serious  matter,  and  should, 
indeed,  be  carefully  computed  in  all  cases,  and  its  results  com- 
bined with  the  fixed  load  and  wind  stresses  already  deter- 
mined. It  will  throw  a  considerable  portion  of  the  fixed  load 
in  the  windward  truss  into  the  leeward.  It  will  also  consid- 
erably increase  the  downward  pull  on  the  windward  side  of  the 
anchorage,  and  relieve  that  at  the  leeward  side  by  the  same 
amount.  These  changes  in  the  downward  pull  of  one  side  of 
the  anchorage  can  be  found  by  dividing  the  wind  moment 
about  the  top  of  the  pier  between  the  anchor  and  cantilever 
arms  by  the  transverse  width  between  the  trusses  where  they 
are  attached  to  the  anchorage.  The  same  effects  occur,  of 
course,  similarly  in  the  anchor  span.  These  overturning 
effects,  resulting  from  the  pressure  of  the  wind  against  the 
structure,  affect  the  fixed-load  stresses  only.  It  may  also  be 
necessary  to  consider  the  truss  stresses  arising  from  the  over- 
turning effect  of  the  wind  on  the  train.     This,  however,  will 


472  APPENDIX    V. 

depend  upon  the  special  circumstances  affecting  any  particu. 
lar  case. 

Art.   5. — Xiconomic  Lengths  of  Spans  and  Arms. 

The  problem  of  the  relative  lengths  of  suspended  spans, 
cantilever  arms,  anchor  arms,  and  anchor  spans  for  the  most 
economical  amounts  of  material  in  them  is  not  capable  of 
exact  mathematical  treatment,  but  may  be  very  simply  solved 
in  a  manner  sufficiently  accurate  for  all  practical  purposes. 
In  order  to  do  this  it  is  only  necessary  to  observe  that  the 
weight  per  lineal  foot  of  any  single  span  of  bridge  structure, 
omitting  the  floor,  is  very  closely  found  by  multiplying  the 
span  length  by  a  numerical  factor,  determined  by  actual  com- 
putation of  weights  of  similar  spans.  In  the  case  of  a  canti- 
lever arm,  the  length  of  that  arm  would  be  used  instead  of 
span  length.     In  the  analysis  which  follows,  there  will  be  used  : 

a  —  cantilever  arm  factor  ;  b  =  suspended  span  factor  ; 

f  =  anchor  "         "     ;  rt^=  anchor  "         " 

L  =  length  of  cantilever  span  —  I  +  2ni  of  Fig.  i,  PI.  XIII. 

/  =  "          suspended  span. 

Economic  Length  of  Suspended  Span  and  Cantilever  A  rm. 

The  total  weight,  W,  of  the  suspended  span  and  cantilever 
arm  trusses  will  be  : 

-{L-lf  +  bP=W, 

Therefore,  by  differentiation : 

a{L-l)-  2bl  =  o.         .-.  /=  — ^  Z.  .     .     .     (i). 

^  ^  a  +  2b  ^  ^ 

Hence,  cantilever  arm  =  w  =  ■ j  L (2)» 

a  +  2b 

If  there  be  taken  loosely,  a  —  20  and  b=  j,  I  will  be  nearly 
0.6  L  and  in  nearly  0.2  L.  It  has  been  found,  however,  that 
/  may  be  advantageously  taken  about  0.5  Z  to  0.55  Z,  and  m 
about  0.25  L. 


CANTILEVERS.  473 

Economic  Length  of  Anchor  Arm, 

There  is  to  be  found  in  this  case  the  entire  weight  of  the 
trusses  for  two  anchor  arms  and  the  cantilever  opening,  or 
the  weight  of  /  +  2n  +  2m  =  5  of  Fig.  i,  PL  XIII.  By  using 
the  proportions  established  in  the  preceding  case,  there  will 
result : 

Anchor  arm  length  =  n. 

Suspended  span  length  = -j  (S  —  2n)  =  I, 

Cantilever  arm  length  =  —         {S  —  2n)  =  tn. 

Hence  the  entire  weight  desired  will  be  : 
ad 


2cn' 


a  + 


-^{S-2ny=W.     ....    (3). 


By  differentiation : 

; r  {S  —  211)  =  O.        .-.   n  =  -7 7- ;  S.        (4). 

a-\-2b^  '  c{a  +  2d}  +  2ab         ^^' 


en  — 


,;l  =  —. TT r  S\  and,  m  —  —. r- -.  S.    (5). 

c{a  +  2b) -\- 2ab  c  {a  ■\- 2b) -\- 2ab         ^^' 

If,  as  before,  there  be  taken  loosely : 

a  =  20\  b  —  7  \  and  c  =  10, 
there  will  result : 

7t  =  0.226  S  ;  w  =  0.1 13  S,  and  /  =  0.32  S. 

It  has  been  found  in  the  best  practice  that  the  anchor  arm 
should  be  1.67  to  2  times  the  length. of  the  cantilever  arm, 
with  the  suspended  span  about  twice  the  length  of  th^  latter, 
or  a  very  little  more. 

Economic  Length  of  the  Anchor  Span. 

The  economic  length  of  the  anchor  span  will  be  found  by 
considering  the  total  weight  of  the  trusses  for  an  anchor  span 


474  APPENDIX    V, 

and  an  adjoining  cantilever  span,  or  the  weight  of  «  +  2m.  +  / 
in  Fig.  2,  PI.  XIII. 

The  proportions  established  in  the  preceding  cases  give : 

Anchor  span  length  =  n. 

Cantilever  arm      "     = r  (S  —  n). 

a  +  2d  ' 


Suspended  span    "     = — y  {S  —  n). 

Hence  the  total  weight  desired  is: 

dn^^-^^AS-tif^W, (6> 

a  +  2b^  '  ^  ' 

Differentiation  then  gives : 

f  ,          ab     \           ab      c^                                ab  o  f^\ 

nid+ T )  = 7  S,       ,',  n  =  -j7 Tx— — T  S.  (7). 


a  +  2b  J       a  +  2b    '  d{a  +  2b)  +  ab 

Hence: 

5;  and  m=     ,,     ^      ..    ,      ,  5.  .     (8). 


d{a  +  2b)  +  ab      '  d  {a  +  2b)  +  ab 

If  the  same  rough  factors  as  before  be  taken,  viz.,  a  =  20\ 
b  =  y,  and  d  ■=  12,  there  will  result : 

«  =0.265;    /  =  0.438  5;   and  ?«  =  0.153  5. 

These  results,  however,  make  n  too  small  for  the  ordinary 
requirements  of  navigation  where  such  exist.  The  strict 
economic  requirements  in  such  cases  are,  therefore,  neglected, 
and  n  taken  from  0.3  to  0.4  5.  These  results  show  the 
importance,  however,  of  making  the  anchor  span,  compara- 
tively speaking,  very  short. 

Art.  6. — Wind    Pressure. 

The  great  importance  of  wind  stresses  in  cantilever  struct- 
ures  makes  it   of    much  interest  to  observe  that  the  latest 


CANTILEVERS.  475 

investigations  indicate  a  somewhat  lower  maximum  on  large 
areas  for  the  usual  highest  winds  than  has  hitherto  been 
contemplated.  The  highest  pressure  actually  observed  and 
measured  at  a  very  exposed  point  on  the  site  of  the  Forth 
bridge,  over  a  period  of  six  years,  on  a  surface  of  300  (15  ft. 
by  20  ft.)  square  feet,  was  27  pounds  per  square  foot, 
although  41  pounds  per  square  foot  was  at  the  same  time 
observed  on  a  small  area  of  1.5  square  feet  immediately 
adjacent  to  the  large  one.  A  pressure  of  50  pounds  per 
square  foot,  therefore,  over  the  entire  surfaces  exposed  in  a 
500  feet  cantilever  span  is  probably  as  rare  at  any  given 
location  as  a  cylone  at  the  same  place. 

Again,  some  later  investigations  of  a  most  valuable  char- 
acter, by  Mr.  O.  T.  Crosby,  and  given  in  his  paper,  "  An 
Experimental  Study  of  Atmospheric  Resistance,"  read  before 
the  West  Point  Branch  of  the  United  States  Military  Service 
Institution,  in  1890,  appear  to  quite  invalidate  the  formula, 

V-      .  .  . 

/*=——,   in   which /*  is  the  pressure  in  pounds   per  square 

foot,  and  Fthe  wind  velocity  in  miles  per  hour,  given  in 
Art.  80,  page  370.  Mr.  Crosby's  experiments  with  a  surface 
one  square  foot  in  area  show  that  with  velocities  varying 
from  30  to  130  miles  per  hour,  very  closely  : 

7 

This  is  such  a  radical  departure  from  the  hitherto  accepted 
values  for  P  that  further  tests  are  desirable,  although  it  is 
difificult  to  detect  any  ground  of  error  in  Mr.  Crosby's 
results. 


PIJ 


PI.  I 


Fig.i 

B 

c   . 

J) 

jP 

-^      .   i      . 

.  G^ 

ff" 

w 

h\- 

/\              "'" 

A 

■:• 

/-V'^ 

/        \ 

--A  V 

\ 

/^T   '?y' 

'\ 

--'"/ 

^ 

/              ) 

^ 

7^ 

-'-\a^ 

:3 

yyi' 

a- 

?     -'     ' 

j"-^ 

^.     V 

,.'-''/> - -- 

ir 

:>R 

jr<i^ 

^^_^^ 

M 


Fig.3. 
B   -33.  gr     J>    -55.  ai      F   -gj  oe     H 


"   ' 

/ 

/ 

2 

/ 

J 

/ 

\ 

\ 

- 

r 

+ 

2 

/ 

r 

/ 

CI 

•4- 

/ 

/ 

»/\ 

\ 

\ 

/ 

1 3.1.  87 

/ 

tSS.  91 

/ 

tff  J.  0!& 

/  +6s.oe 

\ 

\ 

A 

1 

1 

6 

2 

i 

3 

6 

+ 

/I 

p 

r 

t* 

FiQ4- 


^     /..     B 


I 'in. 


^ 


ri.m. 


r  n 


o'  p'  q'  t" 


o   c  f-^  o 

o    0  v.,  o 


A 


a 

b 

^ 

;:      "" 

T 

T' 

^(1 

li 

V 

^ 

1  j 

^ 

■f'ff 

^•L' 

E 


Ua     U 


«7^ 


■^- 


E 


Fiff.C. 


"A 


Pi.m. 


V. 


Fi().  i 


a 

h 

W 

\\ — 

-J 

r 

\f^ 

■5\ 

-^- —I  X-Fiq.G. 


/y./i: 


PiJV. 


/v.\ 


P/.V. 


n.Yf. 


Pi.rr. 


pi,m. 


pi.m. 


/y.  17//. 


/'i.vm. 


b,       J 


/  h.  J^\ 


/y./i. 


j_ 


c 


/'/.X. 


I^'      G     H     K     L      O     r      Q       S 


m.  5 


/v.x. 


\  ( 


Fin.    I. 


C 


/v.  XI 


W 


/6 


c 


o  o  o   o  o 


o    o  o   o   o 


~^ 


1 


FitfJf 


0,         ;i 

>->'                o 
0.                !o 

o'                '  o 

"i — 

/ 

/'///.  /'T. 


Ol                   lO 

0'           ;o 

^ 

K 

i^ 

■6 

i 

Ol                  |0 
O,                  lO 
C,                 10 

Oi                  10 

1 
1 

/ 

/;./.  /^. 


1-  /2.' 


T'/. 


4"ip  tx/^ovf^^  fY/ul  de^low 


Fiq.  9. 


ir. 


I'l.  .\l 


„  ./         K       ff         JT        f      -  •^■ 


♦  «£?=-    i. 


^0j; 


oi  *^  is 


iO:^ 


Fi<t./f 


\f'll/.  /o 


t 


o:^ 

'  \ 

o 

'•;' 

1 

\ 

o 

0 

i 

o      o 

:f    1 

■\    o 

^5 — 

3" 

Fuj  .  /„ 


F"/.  0 


e 


r 


I'l  MI. 


Fig.  4, 


,    , U-^j^ 


Fig.JZ. 


u 


^p 


^  h 


Q^  9.  P 


Fii^J9. 


Surface  of  f  •>Roc}:     uJ 


I'l  xu. 


/; 

/ 

o     !- — 

Fiff.f. 

T: 

/■■ 

B\\0       \       II-  fiff-^ 


-' 

Fiff.J. 

.. 

(  o 

V 

.1 

D 

O   ) 

v_c.^ 

' 

--i-' 

ftrf.  7. 


®)     c 


/><7./. 


3,0tt3 

•H 

^^ 

"fe 

Fiff.W. 

H 1 

II             : 

/>y./Z 


te:'oi  o  To;  O'Ol 


-...I3"i--. 

Fig.n. 


Fig.  12. 


^^ 


Kjr./^ 


■^^ 


//y./.9. 


Pf.X/Il. 


AZ- 


D 


E 


pi.xm. 


IXIXIXiXiXlXlXlXlXIXlXlXIXIXIXIXIXIXIx:^!^ 

A                                        B                     \C  „.._/?           E 

I  FiQ.  5. 

,.... n  _^_       m_ , I, ^ 

A        e        r. a, B                 C  D 


ZC 


R'R" 
-Ft, 


Fig.  4 


SHORT-TITLE     CATALOGUE 

OP  THE 

PUBLICATIONS 

OF 

JOHN   WILET   &    SONS, 

New  York. 
London:   CHAPMAN  &  HALL,  Limitbd. 


ARRANGED  UNDER  SUBJECTS. 


Descriptive  circulars  sent  on  application.  Rooks  marked  with  an  asterisk  (*)  are  sold 
at  net  prices  only,  a  double  asterisk  (**)  books  sold  under  the  rules  of  the  American 
Publishers'  Association  at  net  prices  subject  to  an  extra  charge  for  postage.  All  books 
are  bound  in  cloth  unless  otherwise   stated. 


AGRICULTURE. 

Armsby's  Manual  of  Cattle-feeding lamo,  Si  75 

Principles  of  Animal  Nutrition .8vo,  4  00 

Budd  and  Hansen's  American  Horticultural  Manual: 

Part  I.  Propagation,  Culture,  and  Improvement i2mo,  i  50 

Part  II.  Systematic  Pomology 1211:0,  i  50 

Downlng's  Fruits  and  Fruit-trees  of  America _. 8vo,  5  00 

Elliott's  Engineering  for  Land  Drainage i2mo,  1  50 

Practical  Farm  Drainage izmo,  i  00 

Green's  Principles  of  American  Forestry i2mo,  i  50 

Grotenfelt's  Principles  of  Modern  Dairy  Practice.     (WoU.) i2mo,  2  00 

Kemp's  Landscape  Gardening i2mo,  2  50 

Maynard's  Landscape  Gardening  as  Applied  to  Home  Decoration i2mo,  1  qo 

Sanderson's  Insects  Injurious  to  Staple  Crops i2mo,  i  50 

Insects  Injurious  to  Garden  Crops.     (In  preparation.) 

Insects  Injuring  Fruits.     (In  preparation.) 

Stockbridge's  Rocks  and  Soils 8vo,  2  50 

Woll's  Handbook  for  Farmers  and  Dairymen i6mo,  i  50 

ARCHITECTURE. 

Baldwin's  Steam  Heating  for  Buildings i2mo,  2  50 

Bashore's  Sanitation  of  a  Country  House i2mo,  i   00 

Berg's  Buildings  and  Structures  of  American  Railroads 4to,  5  00 

Birkmire's  Planning  and  Construction  of  American  Theatres Svo,  3  00 

Architectural  Iron  and  Steel Svo,  3  50 

Compound  Riveted  Girders  as  Applied  in  Buildings Svo,  2  00 

Planning  and  Construction  of  High  Office  Buildings Svo  3  30 

Skeleton  Construction  in  Buildings Svo,  3  00 

Brigg's  Modern  American  School  Buildings Svo,  4  00 

Carpenter's  Heating  and  Ventilating  of  Buildings Svo.  4  00 

Freitag's  Architectural  Engineering.- Svo,  3  50 

Fireproofing  of  Steel  Buildings Svo,  2  50 

French  and  Ives's  Stereotomy Svo,  2  50 

Gerhard's  Guide  to  Sanitary  House-inspection i6mo,  1  00 

Theatre  Fires  and  Panics.     i2mo,  1  50 

Holly's  Carpenters'  and  Joiners'  Handbook iSmo,  75 

Tohnson's  Statics  by  Algebraic  and  Graphic  Methods Svo,  2  oe 

1 


4 

oo 

4 

oo 

5 

oo 

7 

50 

4 

00 

3 

00 

I 

50 

3 

50 

2 

3 

3 

00 

6 

00 

6 

50 

5 

00 

5 

50 

3 

00 

4 

00 

2 

50 

Kidder's  Architects' and  Builders' Pocket-book.  Rewritten  Edition.  i6mo,mor.,  5  00 

Merrill's  Stones  for  Building  and  Decoration 8vo,    5  00 

Non-metallic  Minerals:   Their  Occurrence  and  Uses 8vo, 

Monckton's  Stair-building 4to, 

Patton's  Practical  Treatise  on  Foundations 8vo, 

Peabody's  Naval  Architecture Svo, 

Richey's  Handbook  for  Superintendents  of  Construction i6mo,  mor  , 

Sabin's  Industrial  and  Artistic  Technology  of  Paints  and  Varnish Svo, 

Siebert  and  Biggin's  Modem  Stone-cutting  and  Masonry Svo, 

Snow's  Principal  Species  of  Wood Svo, 

Sondericker's  Graphic  Statics  with  Applications  to  Trusses,  Beams,  and  Arches. 

Svo, 

Towne's  Locks  and  Builders*  Hardware iSmo,  morocco. 

Wait's  Engineering  and  Architectural  Jurisprudence Svo, 

Sheep, 
Law  of  Operations  Preliminary  to  Construction  in  Engineering  and  Archi- 
tecture  Svo, 

Sheep, 

Law  of  Contracts 8vo, 

Wood's  Rustless  Coatings:   Corrosion  and  Electrolysis  of  Iron  and  Steel.  .Svo, 

Woodbury's  F're  Protection  of  Mills Svo, 

Worcester  and  Atkinson's  Small  Hospitals,  Establishment  and  Maintenance, 
Suggestions  for  Hospital  Architecture,  with  Flans  for  a  Small  Hospital 

i2mo,     I   25 
The  World's  Columbian  Exposition  of  1S93 ,. Large  4to,     1  00 

ARMY  AND  NAVY. 

Bernadou's  Smokeless  Powder,  Nitro-cellulose,  and  the  Theory  of  the  Cellulose 
Molecule 1 2mo, 

*  Bruff's  Text-boek  Ordnance  and  Gunnery Svo, 

Chase's  Screw  Propellers  and  lUarine  iropuk'cv Svo, 

Cloke's  Gunner's    Examiner Svo, 

Crarg's  Azimuth 4to, 

Crehore  and  Squier's  Polarizing  Photo-chronograph Svo, 

Cronkhite's  Gunnery  for  Non-commissioned  Officers 24mo,  morocco, 

*  Davis's  Elements  of  Law Svo, 

*  Treatise  on  the  Military  Law  of  United  States Svo, 

Sheep, 

De  Brack's  Cavalry  Outposts  Duties.      (Carr.) 24mo,  morocco,  2  00 

Dietz's  Soldier's  First  Aid  Handbook i6mo,  morocco,  i  25 

*  Dredge's  Modern  French  Artillery 4to,  half  morocco,  15  00  • 

Durand's  Resistance  and  Propulsion  of  Ships Svo,  5  00 

*  Dyer's  Handbook  of  Light  Artillery i2mo,  3  00 

Eissler's  Modern  High  Explosives Svo,  4  00 

*  Fiebeger's  Text-book  on  Field  Fortification Small  Svo,  2  00 

Hamilton's  The  Gunner's  Catechism iSmo,  i  00 

*  Hoff's  Elementary  Naval  Tactics Svo,  i  50 

Ingalls's  Handbook  of  Problems  in  Direct  Fire Svo,  4  00 

*  Ballistic  Tables Svo,     i  50 

^  Lyons's  Treatise  on  Electromagnetic  Phenomena.  Vols.  I.  and  H. .  Svo,  each,    6  00 

*  Mahan's  Permanent  Fortifications.    (Mercur.) Svo,  half  morocco,     7  50 

Manual  for  Courts-martial i6mo,  morocco,     1  50 

*  Mercur's  Attack  of  Fortified  Places i2mo,    2  00 

*  Elements  of  the  Art  of  War Svo,  4  00 

Metcalf's  Cost  of  Manufactiires — And  the  Administration  of  Workshops.  .Svo,  5  00 

*  Ordnance  and  Gunnery.     2  vols i2mo,  5  00 

Murray's  Infantry  Drill  Regulations iSmo,  paper,  10 

Fison's  Adjutants'  Manual 24mo,  1  00 

Peabody's  Naval  Architecture Svo,  7  50 

3 


2 

50 

6 

00 

3 

00 

I 

50 

3 

50 

3 

00 

2 

00 

2 

50 

7 

00 

7 

50 

3 

oo 

3 

00 

1 

00 

2 

00 

3 

oo 

3 

oo 

I 

50 

I 

SO 

2 

50 

3 

50 

4 

00 

*  i-helps's  Practical  Marine  Surveying 8vo,  2  50 

Powell's  Army  Officer's  Examiner i2mo,  4  00 

Sharpe's  Art  of  Subsisting  Armies  in  War i8mo.  morocco,  i  50 

*  Walke's  Lectures  on  Explosives 8vo,  4  00 

*  Wheeler's  Siege  Operations  and  Military  Mining 8vo,  2  00 

Winthrop's  Abridgment  of  Military  Law i2mo,  2  50 

WoodhuU's  Notes  on  Military  Hygiene i6mo,  i  50 

Young's  Simple  Elements  of  Navigation i6mo,  morocco,  i  00 

Second  Edition,  Enlarged  and  Revised i6mo,  morocco,  2  00 

ASSAYING. 

Fletcher's  Practical  Instructions  i*.  Quantitative  Assaying  with  the  Blowpipe. 

i2mo,  morocco,     i  50 

Furman's  Manual  of  Practical  Assaying Svo, 

Lodge's  Notes  on  Assaying  and  Metallurgical  Laboratory  Experiments.  .  .   Svo, 

Miller's  Manual  of  Assaying l2mo, 

O'DriscoU's  Notes  on  the  Treatment  of  Gold  Ores Svo, 

Ricketts  and  Miller's  Notes  on  Assaying Svo, 

Ulke's  Modern  Electrolytic  Copper  Refining Svo, 

Wilson's  Cyanide  Processes i2mo, 

Chlorination  Process i2mo, 

ASTRONOMY. 

Comstock's  Field  Astronomy  for  Engineers Svo, 

Craig's  Azimuth 4to, 

Doolittle's  Treatise  on  Practical  Astronomy Svo, 

Gore's  Elements  of  Geodesy Svo,    2  50 

Hayford's  Text-book  of  Geodetic  Astronomy Svo,    3  00 

Merriman's  Elements  of  Precise  Surveying  and  Geodesy Svo,    2  50 

*  Michie  and  Harlow's  Practical  Astronomy Svo,    3  00 

*  White's  Elements  of  Theoretical  and  Descriptive  Astronomy i2mo,    2  00 

BOTANY. 

Davenport's  Statistical  Methods,  with  Special  Reierence  to  Biological  Variation. 

i6mo,  morocco,     1  25 

Thome'  and  Bennett's  Structural  and  Physiological  Botany i6mo,    2  25 

Westermaier's  Compendium  of  General  Botany.     (Schneider.) Svo,    2  00 

CHEMISTRY. 

Adriance's  Laboratory  Calculations  and  Specific  Gravity  Tables i2mo, 

Allen's  Tables  for  Iron  Analysis Svo, 

Arnold's  Compendium  of  Chemistry.     (Mandel.) Small  Svo, 

Austen's  Notes  for  Chemical  Students lamo, 

Bernadou's  Smokeless  Powder. — Nitro-cellulose,  and  Theory  of  the  Cellulose 

Molecule i2mo, 

Bolton's  Quantitative  Analysis Svo, 

*  Browning's  Introduction  to  the  Rarer  Elements Svo, 

Brush  and  Penfield's  Manual  of  Determinative  Mineralogy Svo, 

Classen's  Quantitative  Chemical  Analysis  by  Electrolysis.    (Boltwood. ).   Svo, 
Cohn's  Indicators  and  Test-papers i2mo,    2  00 

Tests  and  Reagents *5vo,  3  00 

Crafts's  Short  Course  in  Qualitative  Chemical  Analysis.   (Schaeffer.).  .  .  i2mo,  i  50 
Dolezalek's   Theory  of  the   Lead  Accumulator   (Storage   Battery).        (Von 

Ende.) i2mo,  2  50 

Drechsel's  Chemical  Reactions.     (Merrill.) i2mo,  i  25 

Duhem's  Thermodynamics  and  Chemistry.     (Burgess.) Svo.  4  00 

Eissler's  Modern  High  Explosives 8vo,  4  00 

Effront's  Enzymes  and  their  Applications.     (Prescott.) 8vo,  3  00 

Erdmann's  Introduction  to  Chemical  Preparations.     (Dixnlap.) i2mo,  i  25 

3 


3 

00 

3 

50 

I 

50 

2 

50 

I 

50 

I 

50 

4 

00 

3 

00 

Fletcher's  Practical  Instructions  in  Quantitative  Assaying  with  the  Blcv  pipe. 

i2mo,  morocco,    i  50 

Fowler's  Sewage  Works  Analyses i2mo,    2  00 

Fresenius's  Manual  of  Qualitative  Chemical  Analysis.     (Wells.) 8vo,    5  00 

Manual  of  Qualitative  Chemical  Analysis.  Part  I.  Descriptive.  (Wells.)  8vo,    3  00 
System   of    Instruction    in    Quantitative    Chemical   Analysis.      (Cchn.) 

2  vols 8vo,  1 2  50 

Fuertes's  Water  and  Public  Health i2mo,    1  50 

Flyman's  Manual  of  Practical  Assaying 8vo,    3  00 

*  Getman's  Exercises  in  Physical  Chemistry lamo,    2  00 

Gill's  Gas  and  Fuel  Analysis  for  Engineers i2mo,     i   25 

Grotenfelt's  Principles  of  Modern  Dairy  Practice.     (Woll.) i2mo,    2  00 

Hammarsten's  Text-book  of  Physiological  Chemistry.     (Mandel.) 8vo,    4  00 

Helm's  Principles  of  Mathematical  Chemistry.     (Morgan.) i2nio,     1  50 

Bering's  Ready  Reference  lables  (Conversion  Factors) i6mo  morocco,    2  50 

Hind's  Inorganic  Chemistry 8vo,    3  00 

*  Laboratory  Manual  for  Students i2mo,     1   00 

HoUeman's  Text-book  of  Inorganic  Chemistry.     (Cooper.) 8vo,    2  50 

Text-book  of  Organic  Chemistry.     (Walker  and  Mott.) 8vo,    2  50 

*  Laboratory  Manual  of  Organic  Chemistry.     (Walker.) i2mo,     i  00 

Hopkins's  Oil-chemists'  Handbook 8vo,    3  00 

Jackson's  Directions  for  Laboratory  Work  in  Physiological  Chemistry.  .8vo,     i  25 

Keep's  Cast  Iron 8vo,    2  50 

Ladd's  Manual  of  Quantitative  Chemical  Analysis .- i2mo,     1  00 

Landauer's  Spectrum  Analysis.     (Tingle.) 8vo,    3  00 

*  Langworthy   and   Austen.        The   Occurrence   of  Aluminiurri  in  Vege  able 

Products,  Animal  Products,  and  Natural  Waters 8vc,    2  00 

Lassar-Cohn's  Practical  Urinary  Analysis.     (Lorenz.) i2mo,     i  00 

Application  of  Some   General   Reactions   to   Investigations  in  Organic 

Chemistry.     (Tingle.) i2mo,     i  00 

Leach's  The  Inspection  and  Analysis  of  Food  with  Special  Reference  to  State 

Control 8vo,     7  50 

Lob's  Electrolysis  and  Electrosynthesis  of  Organic  Compounds.  (Lorenz. ).i2mo,  i  00 
Lodge's  Notes  on  Assaying  and  Metallurgical  Laboratory  Experiments.  ..  .8vo,    3  00 

Lunge's  Techno-chemical  Analysis.     (Cohn.) i2mo,     i  00 

Mandel's  Handbook  for  Bio-chemical  Laboratory i2mo,     i  50 

*  Martin's  Laboratory  Guide  to  Qualitative  Analysis  with  the  Blowpipe .  .  i2mo,  60 
Mason's  Water-supply.     (Considered  Principally  from  a  Sanitary  Standpoint.) 

3d  Edition,  Rewritten 8vo,    4  00 

Examination  of  Water.     (Chemical  and  Bacteriological.) i2mo,    i  25 

Matthew's  The  Textile  Fibres • 8vo,    3  50 

Meyer's  Determination  of  Radicles  in  Carbon  Compounds.     (Tingle.).  .i2mo,     i  00 

Miller's  Manual  of  Assaying i2mo,    1  00 

Mixter's  Elementary  Text-book  of  Chemistry i2mo,     r  50 

Morgan's  Outline  of  Theory  of  Solution  and  its  Results i2mo,    i  00 

Elements  of  Physical  Chemistry i2mo,    2  00 

Morse's  Calculations  used  in  Cane-sugar  Factories T6mo,  morocco,     i  50 

Mulliken's  General  Method  for  the  Identification  of  Pure  Organic  Compounds. 

Vol.  I Large  8vo,    5  00 

O'Brine's  Laboratory  Guide  in  Chemical  Analysis 8vo,    2  00 

O'DriscoU's  Notes  on  the  Treatment  of  Gold  Ores 8vo,    2  00 

Cstwald'p  Conversations  on  Chemistry.     Part  One.     (Ramsey.) i2mo,     i  50 

Ostwald's  Conversations  on  Chemistry.     Part  Two.     (Turnbull  ).     (In  Press.) 

*  Penfield's  Notes  on  Determinative  Mineralogy  and  Record  of  Mireral  Tests. 

8vo,  paper,         50 

Pictet's  The  Alkaloids  and  their  Chemical  Constitulicn.     (Biddle.) 8vo,    5  00 

Pinner's  Introduction  to  Organic  Chemistry.     (Austen.) i2mo,    i  50 

Poole's  Calorific  Power  of  Fuels 8vo,    3  00 

Prescott  and  Winslow's  Elements  of  Water  Bacteriology,  with  Special  FcUr- 

ence  to  Sanitary  Water  Analysis i2mo,    i  25 

4 


*  Reisig's  Guide  to  Piece-dyeing 8vo,  25  00 

Richards  and  Woodman's  Air,  Water,  and  Food  from  a  Sanitary  Standpoint  8vq,  3  00 

Richards's  Cost  of  Living  as  Modified  by  Sanitary  Science i2mo,  i  00 

Cost  of  Food,  a  Study  in  Dietaries i2mo,  i  00 

*  Richards  and  Williams's  The  Dietary  Computer 8vo,  i  50 

Ricketts  and  Russell's  Skeleton  Notes  upon  Inorganic   Chemistry.     (Part  I. 

Non-metallic  Elements.) 8vo,  morocco,  73 

Ricketts  and  Miller's  Notes  on  Assaying gvo,  3  00 

Rideal's  Sewage  and  the  Bacterial  Purification  of  Sewage 8vo,  3  50 

Disinfection  and  the  Preservation  of  Food 8vo,  4  00 

Rigg's  Elementary  Manual  for  the  Chemical  Laboratory 8vo,  i  25 

Rostoski's  Serum  Diagnosis.     (Bolduan.) i2iro,  i  00 

Ruddiman's  IncompatibiUties  in  Prescriptions 8vo,  2  00 

Sabin's  Industrial  and  Artistic  Technology  of  Paints  and  Varnish 8vo,  3  00 

Salkowski's  Physiological  and  Pathological  Chemistry.     (Orndorff.) 8vo,  2  50 

Schimpf's  Text-book  of  Volumetric  Analysis i2mo,  2  50 

Essentials  of  Volumetric  Analysis i2mo,  i  25 

Spencer's  Handbook  for  Chemists  of  Beet-sugar  Houses i6mo,  mcrocco,  3  00 

Handbook  for  Sugar  Manufacturers  and  their  Chemists.  .  i6mo,  morocco,  2  00 

Stockbridge's  Rocks  and  Soils 8vo,  2  50 

*  Tillman's  Elementary  Lessons  in  Heat 8vo,  i  50 

*  Descriptive  General  Chemistry 8vo,  3  00 

Treadwell's  Qualitative  Analysis.     (Hall.) 8vo,  3  00 

Quantitative  Analysis.     (Hall.) 8vo,  4  00 

Turneaure  and  Russell's  Public  Water-supplies 8vo,  5  00 

Van  Deventer's  Physical  Chemistry  for  Beginners.     (Boltwood.) i2mo,  i  50 

*  Walke's  Lectures  on  Explosives 8'-o,  4  00 

Washington's  Manual  of  the  Chemical  Analysis  of  Rocks 8-0,  2  00 

Wassermann's  Immune  Sera :  Haemolysins,  Cytotoxins,  and  Precipitins.    (Bol- 
duan.)   ' i2mo,  I  00 

Well's  Laboratory  Guide  in  Qualitative  Chemical  Analysis 8vo,  i  50 

Short  Course  in  Inorganic  Qualitative  Chemical  Analysis  for  Engineerirg 

Students i2mo,  i  50 

Text-book  of  Chemical  Arithmetic i2mo,  i  25 

Whipple's  Microscopy  of  Drinking-water 8vo,  3  50 

Wilson's  Cyanide  Processes i2mo,  i  50 

Chlorination  Process i2mo,  i  50 

Wulling's    Elementary    Course    in  Inorganic,  Pharmaceutical,  and  Medical 

Chemistry 1 2mo,  2  00 

CIVIL  ENGINEERING. 
BRIDGES    AND    ROOFS.       HYDRAULICS.       MATERIALS    OF    ENGINEERING. 
RAILWAY  ENGINEERING. 

Baker's  Engineers'  Surveying  Instruments i2mo,  3  00 

Bixby's  Graphical  Computing  Table Paper  19+  ■;  24}  inches.  25 

**  Burr's  Ancient  and  Modern  Engineering  and  the  Isthmian  Canal.     (Postage, 

27  cents  additional.) 8vo,  3  50 

Comstock's  Field  Astronomy  for  Engineers 8vo,  2  50 

Davis's  Elevation  and  Stadia  Tables 8vo,  i  00 

Elliott's  Engineering  for  Land  Drainage i2mo,  i  50 

Practical  Farm  Drainage i2mo,  i  00 

*Fiebeger's  Treatise  on  Civil  Engineering 8vo,  5  00 

Folwell's  Sewerage.     (Designing  and  Maintenance.) 8vo,  3  00 

Freitag's  Architectural  Engineering.     2d  Edition,  Rewritten 8vo,  3  50 

French  and  Ives's  Stereotomy 8vo,  2  50 

Goodhue's  Municipal  Improvements i2mo,  i  75 

Goodrich's  Economic  Disposal  of  Towns'  Refuse 8vo,  3  50 

Gore's  Elements  of  Geodesy 8vo,  2  50 

Hayford's  Text-book  of  Geodetic  Astronomy 8vo,  3  00 

Bering's  Ready  Reference  Tables  (Conversion  Factors) i6mo,  morocco.  2  50 

5 


Howe's  Retaining  Walls  for  Earth i2mo,  i  i$ 

Johnson's  (J.  B.)  Theory  and  Practice  of  Surveying Small  8vo,  4  00 

Johnson's  (L.  J.)  Statics  by  Algebraic  and  Graphic  Methods 8vo,  2  00 

Laplace's  Philosophical  Essay  on  Probabilities.    (Truscott  and  Emory.)- i2n:o,  2  00 

Mahan's  Treatise  on  Civil  Engineering.     (1873.)     (Wood.) 8vo,  5  00 

*  Descriptive  Geometry 8vo,  i  50 

Merriman's  Elements  of  Precise  Surveying  and  Geodesy 8vo,  2  50 

Elements  of  Sanitary  Engineering 8vo,  2  00 

Merriman  and  Brooks's  Handbpok  for  Surveyors i6mo,  morocco,  2  00 

Nugent's  Plane  Surveying 8vo,  3  50 

Ogden's  Sewer  Design i2mo,  2  00 

Patten's  Treatise  on  Civil  Engineering Svo  half  leather,  7  50 

Reed's  Topographical  Drawing  and  Sketching 4to,  5  00 

Rideal's  Sewage  and  the  Bacterial  Purification  of  Sewat,a Svo,  3  50 

Siebert  and  Biggin's  Modern  Stone-cutting  and  Masonry Svo,  i  50 

Smith's  Manual  of  Topographical  Drawing.     (McMillan.) Svo,  2  50 

Sondericker's  Graphic  Statics,  with  Applications  to  Trusses,  Beams,  and  Arches. 

Svo,  2  00 

Taylor  and  Thompson's  Treatise  on  Concrete,  Plain  and  Reinforced Svo,  5  00 

*  Trautwine's  Civil  Engineer's  Pocket-book i6mo,  morocco,  5  00 

Wait's  Engineering  and  Archi'ectural  Jurisprudence Svo,  6  00 

Sheep,  6  50 
Law  of  Operations  Preliminary  to  Construction  in  Engineering  and  Archi- 
tecture  Svo,  5  DO 

Sheep,  5  50 

Law  of  Contracts Svo,  3  00 

Warren's  Stereotomy — Problems  in  Stone-cutting Svo,  2  50 

Webb's  Problems  in  the  Use  and  Adjustment  of  Engineering  "Instruments. 

i6mo,  morocco,  i  25 

*  Wheeler  s  Elementary  Course  of  Civil  Engineering Svo,  4  00 

Wilson's  Topographic  Surveying Svo,  3  50 

BRIDGES  AND  ROOFS. 

Boiler's  Practical  Treatise  on  the  Construction  of  Iron  Highway  Bridges.   Sro,  2  00 

*  Thames  River  Bridge 4to,  paper,  5  00 

Burr's  Course  on  the  Stresses  in  Bridges  and  Roof  Trusses,  Arched  Ribs,  and 

Suspension  Bridges Svo,  3  50 

Burr  and  Falk's  Influence  Lines  for  Bridge  and  Roof  Computations.  .  .   Svo,  3  00 

Du  Bois's  Mechanics  of  Engineering.     Vol.  II Small  4to,  10  00 

Foster's  Treatise  on  Wooden  Trestle  Bridges 4to,  5  00 

Fowler's  Ordinary  Foundations Svo,  3  5p 

Greene's  Roof  Trusses Svo,  i  25 

Bridge  Trusses Svo,  2  50 

Arches  in  Wood,  Iron,  and  Stone Svo,  2  50 

Howe's  Treatise  on  Arches Svo,  4  00 

Design  of  fi.mple  Roof-trusses  in  Wood  and  Steel Svo,  2  00 

Johnson,  Bryan,  and  Tumeaure's  Theory  and  Practice  in  the  Designing  of 

Modern  Framed  Structures Small  4to,  10  00 

Merriman  and  Jacoby's  Text-book  on  Roofs  and  Bridges: 

Part  I.     Stresses  in  Simple  Trusses Svo,  2  50 

Part  11.     Graphic  Statics Svo,  2  50 

Part  in.     Bridge  Design Svo,  2  50 

Part  rV.     Higher  Structures Svo,  2  50 

Morison's  Memphis  Bridge 4to,  10  00 

Waddell's  De  Pontibus,  a  Pocket-book  for  Bridge  Engineers.  .  i6mo,  morocco,  3  00 

Specifications  for  Steel  Bridges i2mo,  1  25 

Wood's  Treatise  on  the  Theory  of  the  Construction  of  Bridges  and  Roofs .  .  Svo,  2  OO 
Wright's  Designing  of  Draw-spans: 

Part  I.     Plate-girder  Draws Svo,  2  50 

Part  n.     Riveted-truss  and  Pin-connected  Long-span  Draws Svo,  2  50 

Two  parts  in  one  volume Svo,  3  50 

6 


HYDRAULICS. 

Bazin's  Experiments  upon  the  Contraction  of  the  Liquid  Vein  Issuing  from 

an  Orifice.     (Trautwine.) 8vo,  2  00 

Bovey's  Treatise  on  Hydraulics 8vo,  5  00 

Church's  Mechanics  of  Engineering 8vo,  6  00 

Diagrams  of  Mean  Velocity  of  Water  in  Open  Channels paper,  1  50 

Coffin's  Graphical  Solution  of  Hydraulic  Problems i6mo,  morocco,  2  50 

Flather's  Dynamometers,  and  the  Measurement  of  Power i2mo,  3  00 

Folwell's  Water-supply  Engineering 8vo,  4  00 

Frizell's  Water-power. 8vo,  5  00 

Fuertes's  Water  and  Public  Health i2mo,  i   50 

Water-filtration  Works i2mo,  2  50 

Ganguillet  and  Kutter's  General  Formula  for  the  Uniform  Flow  of  Water  in 

Rivers  and  Other  Channels.     (Hering  and  Trautwine.) 8vo,  4  00 

Hazen's  Filtration  of  Public  Water-supply 8vo,  3  00 

Hazlehurst's  Towers  and  Tanks  for  Water-works 8vo,  2  50 

Herschel's  115  Experiments  on  the  Carrying  Capacity  of  Large,  Riveted,  Metal 

Conduits 8vo,  2  00 

Mason's  Water-supply.     (Considered  Principally  from  a  Sanitary  Standpoint.) 

8vo,  4  00 

Merriman's  Treatise  on  Hydraulics 8vo,  5  00 

*  Michie's  Elements  of  Analytical  Mechanics 8vo,  4  00 

Schuyler's   Reservoirs   for   Irrigation,   Water-power,   and   Domestic   Water- 
supply Large  8vo,  s  00 

**  Thomas  and  Watt's  Improvement  of  Rivers.     (Post.,  44c.  additional.  ).4to,  6  00 

Turneaure  and  Russell's  Public  Water-supplies Svo,  5  00 

Wegmann's  Design  and  Construction  of  Dams 4to,  5  00 

Water-supply  of  the  City  of  New  York  from  1658  to  1895 4to,  10  00 

Williams  and  Hazen's  Hydraulic  Tables Svo,  i  50 

Wilson's  Irrigation  Engineering * Small  8vo,  4  00 

Wolff's  Windmill  as  a  Prime  Mover 8vo,  3  00 

Wood's  Turbines. Svo,  2  50 

Elements  of  Analytical  Mechanics Svo,  3  00 

MATERIALS  OF  ENGINEERING. 

Baker's  Treatise  on  Masonry  Construction Svo,  5  00 

Roads  and  Pavements Svo,  5  00 

Black's  United  States  Public  Works Oblong  4to,  5  00 

Bovey's  Strength  of  Materials  and  Theory  of  Structures Svo,  7  50 

Burr's  Elasticity  and  Resistance  of  the  Materials  of  Engineering Svo,  7  50 

Byrne's  Highway  Construction Svo,  5  00 

Inspection  of  the  Materials  and  Workmanship  Employed  in  Construction. 

i6mo,  3  00 

Church's  Mechanics  of  Engineering Svo,  6  00 

Du  Bois's  Mechanics  of  Engineering.     Vol.  I Smt^ll  4to,  7  50 

^Eckel's  Cements,  Limes,  and  Plasters Svo,  6  00 

Johnson's  Materials  of  Construction Large  Svo,  6  00 

Fowler's  Ordinary  Foundations Svo,  3  50 

Keep's  Cast  Iron Svo,  2  50 

Lanza's  Applied  Mechanics Svo,  7  50 

Marten's  Handbook  on  Testing  Materials.     (Henning.)     2  vols Svo,  7  50 

Merrill's  Stones  for  Building  and  Decoration Svo,  5  00 

Merriman's  Mechanics  of  Materials.                                 Svo,  5  00 

Strength  of  Materials i2mo,  i  00 

Metcalf's  Steel.     A  Manual  for  Steel-users i2mo,  2  00 

Patton's  Practical  Treatise  on  Foundations Svo,  5  00 

Richardson's  Modern  Asphalt  Pavements Svo,  3  00 

Richey's  Handbook  for  Superintendents  of  Construction i6ino,  mor.,  4  00 

Rockwell's  Roads  and  Pavements  in  France i2mo,  i  25 

7 


Sabin's  Industrial  and  Artistic  Tectinoiogy  of  Paints  and  Varnish 8vo,  3  oo 

Smith's  Materials  of  Machines i2mo,  i  00 

Snow's  Principal  Species  of  Wood 8vo,  3  50 

Spalding's  Hydraulic  Cement lamo,  2  00 

Text-book  on  Roads  and  Pavements i2mo,  2  00 

Taylor  and  Thompson's  Treatise  on  Concrete,  Plain  and  Reinforced 8vo,  5  00 

Thurston's  Materials  of  Engineering.     3  Parts Svo,  8  00 

Part  I.     Non-metallic  Materials  of  Engineering  and  Metallurgy 8vo,  2  00 

Part  n.     Iron  and  Steel 8vo,  3  50 

Part  in.     A  Treatise  on  Brasses,  Bronzes,  and  Other  AUoys  and  their 

Constituents 8vo,  2  50 

Thurston's  Text-book  of  the  Materials  of  Construction 8vo,  5  00 

Tillson's  Street  Pavements  and  Paving  Materials Svo,  4  00 

Waddell's  De  Pontibus.    (*  Pocket-book  for  Bridge  Engineers.).  .  i6mo,  mor.,  3  00 

Specifications  for  StL  i  Bridges i2mo,  i  25 

Wood's  (De  V.)  Treatise  on  the  Resistance  of  Materials,  and  an  Appendix  on 

the  Preservation  of  Timber Svo,  2  00 

Wood's  (De  V.)  Elements  of  Analytical  Mechanics Svo,  3  00 

Wood's  (M.  P.)  Rustless  Coatings:    Corrosion  and  Electrolysis  of  Iron  and 

SteeL Svo,  4  00 

RAILWAY  ENGINEERING. 

Andrew's  Handbook  for  Street  Railway  Engineers 3x5  inches,  morocco,  i  25 

Berg's  Buildings  and  Structures  of  American  Railroads 4to,  5  00 

Brook's  Handbook  of  Street  Railroad  Location i6mo,  morocco,  i  50 

Butt's  Civil  Engineer's  Field-book i6mo,  morocco,  2  50 

Crandall's  Transition  Curve i6mo,  morocco,  i  50 

Railway  and  Other  Earthwork  Tables Svo,  i  50 

Dawson's  "Engineering"  and  Electric  Traction  Pocket-book.  .  i6mo,  morocco,  5  00 

Dredge's  History  of  the  Pennsylvania  Rai'lroad:    (1S79) Paper,  5  00 

*  Drinker's  Tunnelling,  Explosive  Compounds,  and  Rock  Drills. 4to,  half  mor.,  25  00 

Fisher's  Table  of  Cubic  Yards Cardboard,  25 

Godwin's  Railroad  Engineers'  Field-book  and  Explorers'  Guide.  .  .  i6mo,  mor.,  2  50 

Howard's  Transition  Curve  Field-book '.  .  i6mo,  morocco,  i  50 

Hudson's  Tables  for  Calculating  the  Cubic  Contents  of  Excavations  and  Em- 
bankments  Svo,  I  GO 

Mohtor  and  Beard's  Manual  for  Resident  Engineers i6mo,  i  00 

Kagle's  Field  Manual  for  Railroad  Engineers i6mo,  morocco,  3  00 

Philbrick's  Field  Manual  for  Engineers i6mo,  morocco,  3  00 

Searles's  Field  Engineering i6mo,  morocco,  3  00 

Railroad  SpiraL i6mo,  morocco,  i  50 

Taylor's  Prismoidal  Formulae  and  Earthwork Svo,  i  50 

*  Trautwine's  Method  of  Calculating  the  Cube  Contents  of  Excavations  and 

Embankments  by  the  Aid  of  Diagrams Svo,  2  00 

The  Field  Practice  of  Laying  Out  Circular  Curves  for  Railroads. 

i2mo,  morocco,  2  50 

Cross-section  Sheet Paper,  25 

Webb's  Railroad  Construction i6mo,  morocco,  5  00 

WeUington's  Economic  Theory  of  the  Location  of  Railways Small  Svo,  5  00 

DRAWING. 

Barr's  Kinematics  of  Machinery Svo,    2  50 

*  Bartlett's  Mechanical  Drawing Svo,    3  00 

*  "  "  "        Abridged  Ed Svo,  i  50 

Coolidge's  Manual  of  Drawing Svo,  paper  r  00 

Coolidge  and  Freeman's  Elements  of  General  Drafting  for  Mechanical  Engi- 
neers  Oblong  4to,  2  50 

Durley's  Kinematics  of  Machines Svo,    4  00 

Emch's  Introduction  to  Projective  Geometry  and  its  Applications Svo.    2  50 


Hill's  Text-book  on  Shades  and  Shadows,  and  Perspective 8vo,    2  00 

Jamison's  Elements  of  Mechanical  Drawing 8vo,    2  50 

Advanced  Mechanical  Drawing gvo,     2  00 

Jones's  Machine  Design: 

Part  I.     Kinematics  of  Machinery 8vo,     i  50 

Part  II.     Form,  Strength,  and  Proportions  of  Parts 8vo,    3  00 

MacCord's  Elements  of  Descriptive  Geometry 8vo,    3  00 

Kinematics;   or.  Practical  Mechanism 8vo,    5  00 

Mechanical  Drawing 4to,    4  00 

Velocity  Diagrams 8vo,     i  50 

♦  Mahan's  Descriptive  Geometry  and  Stone-cutting 8to,     i  50 

Industrial  Drawing.     (Thompson.) 8vo,    3  50 

Moyer's  Descriptive  Geometry 8vo,     2  00 

Reed's  Topographical  Drawing  and  Sketching 4to,    5  00 

Reid's  Course  in  Mechanical  Drawing 8vo,    2  00 

Text-book  of  Mechanical  Drawing  and  Elementary  Machine  Design. 8vo, 

Robinson's  Principles  of  Mechanism 8vo, 

Schwamb  and  Merrill's  Elements  of  Mechanism 8vo, 

Smith's  Manual  of  Topographical  Drawing.     (McMillan.) 8vo, 

Warren's  Elements  of  Plane  and  Solid  Free-hand  Geometrical  Drawing.  i2mo. 

Drafting  Instruments  and  Operations i2mo. 

Manual  of  Elementary  Projection  Drawing i2mo. 

Manual  of  Elementary  Problems  in  the  Linear  Perspective  of  Form  and 

Shadow i2mo, 

Plane  Problems  in  Elementary  Geometry i2mo, 

Primary  Geometry i2mo, 

Elements  of  Descriptive  Geometry,  Shadows,  and  Perspective. 8vo, 

General  Problems  of  Shades  and  Shadows 8vo, 

Elements  of  Machine  Construction  and  Drawing 8vo, 

Problems,  Theorems,  and  Examples  in  Descriptive  Geometry 8vo, 

Weisbach's  Kinematics  and  Power  of  Transmission.    (Hermann  and  Klein)8vo, 

Whelpley's  Practical  Instruction  in  the  Art  of  Letter  Engraving i2mo, 

Wilson's  (H.  M.)  Topographic  Surveying 8vo, 

Wilson's  (V.  T.)  Free-hand  Perspective 8vo, 

Wilson's  (V.  T. )  Free-hand  Lettering 8vo, 

Woolf's  Elementary  Course  in  Descriptive  Geometry Large  8vo, 


ELECTRICITY  AND  PHYSICS. 

Anthony  and  Brackett's  Text-book  of  Physics.     (Magie.) Small  8vo, 

Anthony's  Lecture-notes  on  the  Theory  of  Electrical  Measurements.  .  .  .i2mo, 
Benjamin's  History  of  Electricity, 8vo, 

Voltaic  Cell 8vo, 

Classen's  Quantitative  Chemical  Analysis  by  Electrolysis.     (Boltwood.).8vo, 

Crehore  and  Squier's  Polarizing  Photo-chronograph 8vo, 

Dawson's  "Engineering"  and  Electric  Traction  Pocket-book.  i6mo,  morocco, 
Dolezalek's    Theory    of    the    Lead   Accumulator    (Storage    Battery).      (Von 

Ende.) i2mo, 

Duhem's  Thermodynamics  and  Chemistry.     (Burgess.) 8vo, 

Flather's  Dynamometers,  and  the  Measurement  of  Power i2m«), 

Gilbert's  De  Magnete.      (Mottelay.) 8vo, 

Hanchett's  Alternating  Currents  Explained i2mo,     i  00 

Bering's  Ready  Reference  Tables  (Conversion  Factors) i6mo,  morocco,    2  50 

Holman's  Precision  of  Measurements 8vo,    2  00 

Telescopic   Mirror-scale  Method,  Adjustments,  and   Tests.  .  .  .Large  8vo,         75 

Kinzbrunner's  Testmg  of  Continuous-Current  Machines 8vo,    2  00 

Landauer's  Spectrum  Analysis.      (Tingle.) 8vo,     3  00 

Le  Chatelien's  High-temperature  Measurements.  (Boudouard — Burgess.)  i2mo      3  00 
Lob's  Electrolysis  and  Electrosynthesis  of  Organic  Compounds.  (Lorenz. )  i2mo,     1  00 

9 


3 

00 

3 

00 

3 

00 

2 

50 

I 

00 

I 

25 

I 

5» 

I 

00 

1 

25 

75 

3 

50 

3 

00 

7 

50 

2 

50 

S 

00 

2 

00 

3 

50 

2 

50 

3 

00 

3 

00 

3 

00 

3 

00 

5 

00 

2 

50 

4 

00 

3 

00 

2 

50 

7 

00 

7 

50 

I 

SO 

6 

00 

6 

SO 

5 

00 

5 

50 

3 

00 

*  Lyons's  Treatise  on  Electromagnetic  Phenomena.   Vols.  I.  and  II.  8vo,  each,  6  00 

*  Michie's  Elements  of  Wave  Motion  Relating  to  Sound  and  Light 8vo,  4  00 

Niaudet's  Elementary  Treatise  on  Electric  Batteries.     (Fishback.) i2mo,  2  50 

*  Rosenberg's  Electrical  Engineering.     (Haldane  Gee — Kinzbrunner.).  .  .8vo,  i  50 

Ryan,  Norris,  and  Hoxie's  Electrical  Machinery.     Vol.  1 8vo,  2  50 

Thurston's  Stationary  Steam-engines 8vo,  2  50 

*  Tillman's  Elementary  Lessons  in  Heat 8vo,  i  50 

Tory  and  Pitcher's  Manual  of  Laboratory  Physics Small  8vo,  2  00 

Ulke's  Modem  Electrolytic  Copper  Refining 8vo,  3  00 

LAW. 

*  Davis's  Elements  of  Law 8vo,  2  50 

*  Treatise  on  the  Military  Law  of  United  States 8vo, 

*  Sheep, 

Manual  for  Courts-martial i6mo,  morocco, 

Wait's  Engineering  and  Architectural  Jurisprudence 8vo, 

Sheep, 
Law  of  Operations  Preliminary  to  Construction  in  Engineering  and  Archi- 
tecture  8vo, 

Sheep, 

Law  of  Contracts 8vo, 

Winthrop's  Abridgment  of  MiUtary  Law i2mo,  2  50 

MANUFACTURES. 

Bernadou's  Smokeless  Powder — Nitro-cellulose  and  Theory  of  the  Cellulose 

Molecule i2mo,  2  50 

BoUand's  Iron  Founder i2mo,  2  50 

"  The  Iron  Founder,"  Supplement i2mo,  2  50 

Encyclopedia  of  Founding  and  Dictionary  of  Foundry  Terms  Used  in  the 

Practice  of  Moulding i2mo,  3  00 

Eissler's  Modern  High  Explosives 8vo,  4  00 

Effront's  Enzymes  and  their  Applications.     (Prescott.) 8vo,  3  00 

Fitzgerald's  Boston  Machinist i2mo,  i  00 

Ford's  Boiler  Making  for  Boiler  Makers i8mo,  i  00 

Hopkin's  Oil-chemists'  Handbook 8vo,  3  00 

Keep's  Cast  Iron 8vo,  2  50 

Leach's  The  Inspection  and  Analysis  of  Food  with  Special  Reference  to  State 

Control Large  8vo,  7  50 

Matthews's  The  Textile  Fibres 8vo,  3  50 

Metcalf's  Steel.     A  Manual  for  Steel-users i2mo,  2  00 

Metcalfe's  Cost  of  Manufactures — And  the  Administration  of  V/orkshops .  8vo,  5  00 

Meyer's  Modern  Locomotive  Construction 4to,  10  00 

Morse's  Calculations  used  in  Cane-sugar  Factories i6mo,  morocco,  i  50 

*  Reisig's  Guide  to  Piece-dyeing 8vo,  25  00 

Sabin's  Industrial  and  Artistic  Technology  of  Paints  and  Varnish 8vo,  3  00 

Smith's  Press-working  of  Metals 8vo,  3  00 

Spalding's  Hydraulic  Cement i2mo,  2  00 

Spencer's  Handbook  for  Chemists  of  Beet-sugar  Houses.    . .  .  i6mo,  morocco,  3  00 

Handbook  for  Sugar  Manufacturers  and  their  Chemists    .  i6mo,  morocco,  2  00 

Taylor  and  Thompson's  Treatise  on  Concrete,  Plain  and  Reinforced 8vo,  5  00 

Thurston's  Manual  of  Steam-boilers,  their  Designs,  Construction  and  Opera- 
tion  8vo,  5  00 

*  Walke's  Lectures  on  Explosives 8vo,  4  00 

Ware's  Manufacture  of  Sugar.     (In  press.) 

West's  American  Foundry  Practice i2mo,  2  50 

Moulder's  Text-book i2mo,  2  50 

10 


Wolff's  Windmill  as  a  Prime  Mover 8vo,    3  o* 

Wood's  Rustless  Coatings:   Corrosion  and  Electrolysis  of  Iron  and  Steel.  .8vo,    4  00 


MATHEMATICS. 

Baker's  Elliptic  Functions 8vo,  i  S» 

*  Bass's  Elements  of  Differential  Calculus i2mo,  4  00 

Briggs's  Elements  of  Plane  Analytic  Geometry i2mo,  1  00 

Compton's  Manual  of  Logarithmic  Computations i2mo,  i  50 

Davis's  Introduction  to  the  Logic  of  Algebra 8vo,  i  50 

*  Dickson's  College  Algebra Large  i2mo,  1  50 

*  Introduction  to  the  Theory  of  Algebraic  Equations Large  12 mo,  i  25 

Emch's  Introduction  to  Projective  Geometry  and  its  Applications 8vo,  2  50 

Halsted's  Elements  of  Geometry 8vo,  i  75 

Elementary  Synthetic  Geometry 8vo,  1  50 

■Rational  Geometry i2mo,  i  75 

*  Johnson's  (J.  B.)  Three-place  Logarithmic  Tables:   Vest-pocket  size. paper,  15 

100  copies  for  5  00 

*  Mounted  on  heavy  cardboard,  8  <  10  inches,  23 

10  copies  for  2  00 

Johnson's  (W.  W.)  Elementary  Treatise  on  Differential  Calculus.  .SmahSvo,  3  00 

Johnson's  (W.  W.)  Elementary  Treatise  on  the  Integral  Calculus. Small  8vo,  i  50 

Johnson's  (W.  W.)  Curve  Tracing  in  Cartesian  Co-ordinates i2mo,  i  oo' 

Johnson's  (W.  W.)  Treatise  on  Ordinary  and  Partial  Differential  Equations. 

Small  8vo,  3  50 

Johnson's  (W.  W.)  Theory  of  Errors  and  the  Method  of  Least  Squares.  i2ma,  i  50 

*  Johnson's  (W.  W.)  Theoretical  Mechanics 1200,  3  00 

Laplace's  Philosophical  Essay  on  Probabilities.     (Truscott  and  Emory.) .  i2mo,  2  00 

*  Ludlow  and  Bass.     Elements  of  Trigonometry  and  Logarithmic  and  Other 

Tables 8vo,  3  00 

Trigonometry  and  Tables  published  separately Each,  2  00 

*  Ludlow's  Logarithmic  and  Trigonometric  Tables 8vo,  i  00 

Maurer's  Technical  Mechanics Svo,  4  00 

Merriman  and  Woodward's  Higher  Mathematics 8vo,  5  00 

Merriman's  Method  of  Least  Squares Svo,  2  00 

Rice  and  Johnson's  Elementary  Treatise  on  the  Differential  Calculus. .  Sm.  8vo,  3  00 

Differential  and  Integral  Calculus.     2  vols,  in  one Small  8vo,  2  50 

Wood's  Elements  of  Co-ordinate  Geometry Svo,  2  00 

Trigonometry:  Analytical,  Plane,  and  Spherical i2mo,  i  00 


MECHANICAL  ENGINEERING. 

MATERIALS  OF  ENGINEERING,  STEAM-ENGINES  AND  BOILERS. 

Bacon's  Forge  Practice i2mo,  i  50 

Baldwin's  Steam  Heating  for  Buildings i2mo,  2  50 

Barr's  Kinematics  of  Machinery Svo,  2  50 

*  Bartlett's  Mechanical  Drawing .' Svo,  3  00 

*  "  "  "        Abridged  Ed Svo,     i  50 

Benjamin's  Wrinkles  and  Recipes i2mo,    2  00 

Carpenter's  Experimental  Engineering Svo,    6  00 

Heating  and  Ventilating  Buildings Svo,    4  00 

Gary's  Smoke  Suppression  in  Plants  using  Bituminous  Coal.     (In  Prepara- 
tion.) 

Clerk's  Gas  and  Oil  Engine Small  Svo,    4  00 

Coolidge's  Manual  of  Drawing Svo,  paper,     i  00 

Coolidge  and  Freeman's  Elements  of  General  Drafting  for  Mechanical  En- 
gineers  Oblong  4to,    2  50 

11 


Cromwell's  Treatise  on  Toothed  Gearing i2mo,  i  so 

Treatise  on  Belts  and  PuUeys i2mo,  i  50 

Durley's  Kineniatics  of  Machines 8vo,  4  00 

Flather's  Dynamometers  and  the  Measurement  of  Power i2mo,  3  00 

Rope  Driving i2mo,  2  00 

Gill's  Gas  and  Fuel  Analysis  for  Engineers i2mo,  i  25 

HaU's  Car  Lubrication i2mo,  i  00 

Bering's  Ready  Reference  Tables  (Conversion  Factors) i6mo,  morocco,  2  50 

Button's  The  Gas  Engine 8vo,  5  00 

Jamison's  Mechanical  Drawing 8vo,  2  50 

Jones's  Machine  Design : 

Part  I.     Kinematics  of  Machinery 8vo,  i  50 

Part  II.     Form,  Strength,  and  Proportions  of  Parts 8vo,  3  00 

Kent's  Mechanical  Engineers'  Pocket-book i6mo,  morocco,  5  00 

Kerr's  Power  and  Power  Transmission 8vo,  2  00 

Leonard's  Machine  Shop,  Tools,  and  Methods 8vo,  4  00 

*Lorenz's  Modern  Refrigerating  Machinery.     (Pope,  Baven,  and  Dean.)    .8vo,  4  00 

MacCord's  Kinematics;   or.  Practical  Mechanism 8vo,  5  00 

Mechanical  Drawing 4to,  4  00 

Velocity  Diagrams 8vo,  i  50 

Mahan's  Industrial  Drawing.     (Thompson.) Svo,  3  50 

Poole  s  Calorific  Power  of  Fuels Svo,  3  00 

Reid's  Course  in  Mechanical  Drawing Svo,  2  00 

Text-book  of  Mechanical  Drawing  and  Elementary  Machine  Design. Svo,  3  00 

Richard's  Compressed  Air i2mo,  i  50 

Robinson's  Principles  of  Mechanism Svo,  3  00 

Schwamb  and  Merrill's  Elements  of  Mechanism Svo,  3  00 

Smith's  Press-working  of  Metals Svo,  3  00 

Thurston's   Treatise    on   Friction  and   Lost   Work   in   Machinery   and    Mill 

Work Svo,  3  00 

Animal  as  a  Machine  and  Prime  Motor,  and  the  Laws  of  Energetics.  1 2mo,  i  00 

Warren's  Elements  of  Machine  Construction  and  Drawing Svo,  7  5a 

Weisbach's    Kinematics    and    the    Power    of    Transmission.     (Berrmann — • 

Klein.) Svo,  5  00 

Machinery  of  Transmission  and  Governors.     (Berrmann — Klein.).  .Svo,  5  00 

Wolff's  WindmiU  as  a  Prime  Mover Svo,  3  00 

Wood's  Turbines Svo,  2  50 


MATERIALS   OF    ENGINEERING. 

Bovey's  Strength  of  Materials  and  Theory  of  Structures Svo,  7  50 

Burr's  Elasticity  and  Resistance  of  the  Materials  of  Engineering.     6th  Edition. 

Reset Svo,  7  50 

Church's  Mechanics  of  Engineering Svo,  6  00 

Johnson's  Materials  of  Construction Svo,  6  00 

Keep's  Cast  Iron Svo,  2  50 

Lanza's  Applied  Mechanics Svo,  7  50 

Martens 's  Bandbook  on  Testing'  Materials.     (Benning.) Svo,  7  50 

Merriman's  Mechanics  of  Materials.  Svo,  5  00 

Strength  of  Materials i2mo,  i  00 

Metcalf's  SteeL     A  manual  for  Steel-users i2mo.  2  00 

Sabin's  Industrial  and  Artistic  Technology  of  Paints  and  Varnish Svo,  3  00 

Smith's  Materials  of  Machines i2mo,  i  00 

Thurston's  Materials  of  Engineering 3  vols.,  Svo,  S  00 

Part  II.     Iron  and  Steel 8vo,  3  50 

Part  III.     A  Treatise  on  Brasses,  Bronzes,  and  Other  Alloys  and  their 

Constituents 8vo,  2  50 

Text-book  of  the  Materials  of  Construction 8^0,  5  00 

12 


r 


Wood's  (De  V.)  Treatise  on  the  Resistance  of  Materials  and  an  Appendix  on 

the  Preseivation  of  Timber 8vo,    2  00 

Wood's  (De  V.)  Elements  of  Analytical  Mechanics 8vo,    3  00 

Wood's  (M.  P.)  Rustless  Coatings:    Corrosion  and  Electrolysis  of  Iron  and 

Ste»l. 8vo,    4  00 


STEAM-ENGINES   AND   BOILERS. 

Berry's  Temperature-entropy  Diagram. i2mo,  i  25 

Carnot's  Reflections  on  the  Motive  Power  of  Heat.     (Thurston.) i2mo,  i  50 

Dawson's  "Engineering"  and  Electric  Traction  Pocket-book.  .  .  .i6mo,  mor.,  5  00 

Ford's  Boiler  Making  for  Boiler  Makers i8mo,  1  00 

Goss's  Locomotive  Sparks 8vo,  2  00 

Hemcnway's  Indicator  Practice  and  Steam-engine  Economy i2mo,  2  00 

Button's  Mechanical  Engineering  of  Power  Plants 8vo,  5  00 

Heat  and  Heat-engines 8vo,  5  00 

Kent's  Steam  boiler  Economy 8vo,  4  00 

Kneass's  Practice  and  Theory  of  the  Injector 8vo,  i  50 

MacCord's  Slide-valves 8vo,  2  00 

Meyer's  Modern  Locomotive  Construction •. 4to,  10  00 

Peabody's  Manual  of  the  Steam-engine  Indicator i2mo,  i  50 

Tables  of  the  Properties  of  Saturated  Steam  and  Other  Vapors 8vo,  i  00 

Thermodynamics  of  the  Steam-engine  and  Other  Heat-engines 8vo,  5  00 

Valve-gears  for  Steam-engines 8vo,  2  50 

Peabody  and  Miller's  Steam-boilers ' 8vo,  4  00 

Pray's  Twenty  Years  v/ith  the  Indicator Large  8vo,  2  50 

Pupin's  Thermodynamics  of  Reversible  Cycles  in  Gases  and  Saturated  Vapors. 

(Osterberg.) i2mo,  1  25 

Reagan's  Locomotives:   Simple   Compound,  and  Electric i2mo,  2  50 

Rontgen's  Principles  of  Thermodynamics.     (Du  Bois.) 8vo,  5  00 

Sinclair's  Locomotive  Engine  Running  and  Management i2mo,  2  00 

Smart's  Handbook  of  Engineering  Laboratory  Practice i2mo,  2  50 

Snow's  Steam-boiler  Practice 8vo,  3  00 

Spangler's  Valve-gears 8vo,  2  50 

Notes  on  Thermodynamics i2mo,  i  00 

Spangler,  Greene,  and  Marshall's  Elements  of  Steam-engineering 8vo,  3  00 

Thurston's  Handy  Tables 8vo,  i  50 

Manual  of  the  Steam-engine 2  vols.,  8vo,  10  00 

Part  I.     History,  Structure,  and  Theory 8vo,  6  00 

Part  U.     Design,  Construction,  and  Operation 8vo,  6  00 

Handbook  of  Engine  and  Boiler  Trials,  and  the  Use  of  the  Indicator  and 

the  Prony  Brake 8vo,  5  00 

Stationary  Steam-engines 8vo,  2  50 

Steam-boiler  Explosions  in  Theory  and  in  Practice i2mo,  i  50 

Manual  of  Steam-boilers,  their  Designs,  Construction,  and  Operation 8vo,  5  00 

Weisbach's  Heat,  Steam,  and  Steam-engines.     (Du  Bois.) 8vo,  5  00 

Whitham's  Steam-engine  Design 8vo,  5  00 

Wilson's  Treatise  on  Steam-boilers.     (Flather.) i6mo,  2  50 

Wood's  Thermodynamics,  Heat  Motors,  and  Refrigerating  Machines.  .  .8vo,  4  00 


MECHANICS  AND  MACHINERY. 

Barr's  Kinematics  of  Machinery 8vo,  2  50 

Bovey's  Strength  of  Materials  and  Theory  of  Structures 8vo,  7  50 

Chase's  The  Art  of  Pattern-making ».  .  .  .  i2mo,  2  50 

Churchts  Mechanics  of  Engineering 8vo,  6  00 

13 


Church's  Notes  and  Examples  in  Mechanics 8vo,  2  00 

Compton's  First  Lessons  in  Metal-working i2mo,  i  50 

Compton  and  De  Groodt's  The  Speed  Lathe i2mo,  1  50 

Cromwell's  Treatise  on  Toothed  Gearing i2mo,  i  50 

Treatise  on  Belts  and  Pulleys i2mo,  i  50 

Dana's  Text-book  of  Elementary  Mechanics  for  Colleges  and  Schools.  .i2mo,  1  50 

Dingey's  Machinery  Pattern  Making i2mo,  2  00 

Dredge's  Record  of  the   Transportation  Exhibits  Building  of  the   World's 

Columbian  Exposition  of  1893 4to  half  mcrocco,  5  00 

Du  Bois's  Elementary  Principles  of  Mechanics: 

Vol.      I.     Kinematics 8vo,  3  50 

Vol.    II.     Statics 8vo,  4  00 

Vol.  III.     Kinetics 8vo,  3  50 

Mechanics  of  Engineering.     Vol.    I Small  4to,  7  50 

Vol.  n Small  4to,  10  00 

Durley's  Kinematics  of  Machines 8vo,  4  00 

Fitzgerald's  Boston  Mac'iinist i6mo,  i  00 

Flather's  Dynamometers,  and  the  Measurement  of  Power i2mo,  3  00 

Rope  Driving i2mo,  2  00 

Goss's  Locomotive  Sparks 8vo,  2  00 

Hall's  Car  Lubrication i2mo,  i  00 

Holly's  Art  of  Saw  Filing i8mo,        75 

James's  Kinematics  of  a  Point  and  the  Rational  Mechanics  of  a  Particle.  Sm.8vo,2  00 

*  Johnson's  (W.  W.)  Theoretical  Mechanics i2mo,  3  00 

Johnson's  (L.  J.)  Statics  by  Graphic  and  Algebraic  Methods 8vo,  2  00 

Jones's  Machine  Design: 

Part    I.     Kinematics  of  Machinery 8vo,  1  50 

Part  U.     Form,  Strength,  and  Proportions  of  Parts 8vc,  3  00 

Kerr's  Power  and  Power  Transmission 8vo,  2  00 

Lanza's  Applied  Mechanics 8v6,  7  50 

Leonard's  Machine  Shop,  Tools,  and  Methods 8vo,  4  00 

*Lorenz's  Modern  Refrigerating  Machinery.      (Pope,  Haven,  and  Dean.)  .8vo,  4  00 

MacCord's  Kmematics;   or,  Practical  Mechanism 8vo,  5  00 

Velocity  Diagrams 8vo,  1  50% 

Maurer's  Technical  Mechanics 8vo,  4  00 

Merriman's  Mechanics  of  Materials 8vo,  5  00 

*  Elements  of  Mechanics i2mo,  i  00 

*  Michie's  Elements  of  Analytical  Mechanics 8vo,  4  00 

Heagan's  Locomotives:   Simple,  Compound,  and  Electric i2mo>  2  50 

Eeid's  Course  in  Mechanical  Drawing 8vo,  2  00 

Text-book  of  Mechanical  Drawing  and  Elementary  Machine  Design .  8vo,  3  00 

Pichards's  Compressed  Air i2mo,  i  50 

Robinson's  Principles  of  Mechanism 8vo,  3  00 

Hyan,  Norris,  and  Hoxie's  Electrical  Machinery.     Vol.  1 8vo,  2  50 

Schwamb  and  Merrill's  Elements  of  Mechanism 8vo,  3  00 

Sinclair's  Locomotive-engine  Running  and  Management i2mo,  2  00 

Smith's  (0.)  Press-working  of  Metals 8vo,  3  00 

Smith's  (A.  W.)  Materials  of  Machines i2mo,  i  00 

Spangler,  Greene,  and  Marshall's  Elements  of  Steam-engineering 8vo,  3  00 

Thurston's  Treatise  on  Friction  and  Lost  ^/ork  in    Machinery  and    MiU 

Work 8vo,  3  00 

Animal  as  a  Machine  and  Prime  Motor,  and  the  Laws  of  Energetics. 

i2mo,  I  00 

Warren's  Elements  of  Machine  Construction  and  Drawing 8vo,  7  50 

Weisbach's  Kinematics  and  Power  of  Transmission.   (Herrmann — Klein.  ).8vo,  5  00 

Machinery  of  Transmission  and  Governors.      (Herrmann — Klein.). 8vo,  5  00 

Wood's  Elements  of  Analytical  Mechanics 8vo,  3  00 

Principles  of  Elementary  Mechanics i2mo,  i  25 

Turbines 8vo,  2  50 

The  World's  Columbian  Esposition  of  1893 4to,  i  00 

14 


METALLURGY. 

Egleston's  Metallurgy  of  Silver,  Gold,  and  Mercury: 

VoL    i.     Silver 8vo.  7  50 

VoL  U.     Gold  and  Mercury 8vo,  7  50 

**  Iles's  Lead-smelting.     (.Postage  9  cents  additional.) i2mo,  2  50 

Keep's  Cast  Iron 8vo,  2  50 

Kunhardt's  Practice  of  Ore  Dressing  in  Europe 8vo,  1  go 

Le  Chatelier's  High-temperature  Measuremepts.  (Boudouard — Burgess.  )i2mo,  3  00 

Metcalf's  Steel.     A  Manual  for  Steel-users     i2mo,  2  00 

Smith's  Materials  of  Machines i2mo,  i  00 

Thurston's  Materials  of  Engineering.     In  Three  Parts 8vo  8  00 

Part    II.     Iron  and  Steel 8vo,  3  50 

Part  in.     A  Treatise  on  Brasses,  Bronzes,  and  Other  Alloys  and  their 

Constituents 8vo,  2  50 

TJlke's  Modern  Electrolytic  Copper  Refining 8vo,  3  00 

MINERALOGY. 

Barringer's  Description  of  Minerals  of  Commercial  Value.    Oblong,  morocco,  2  50 

Boyd's  Resources  of  Southwest  Virginia 8vo,  3  00 

Map  of  Southwest  Virignia. Pocket-book  form.  2  00 

Brush's  Manual  of  Determinative  Mineralogy.     (Penfield.) 8vo,  4  00 

Chester's  Catalogue  of  Minerals 8vo,  paper,  i  00 

Cloth,  1  25 

Dictionary  of  the  Names  of  Minerals Svo,  3  50 

Dana's  System  of  Mineralogy Large  8vo,  half  leather,  12  50 

First  Appendix  to  Dana's  New  "  System  of  Mineralogy." Large  8vo,  i  00 

Text-book  of  Mineralogy Svo,  4  00 

Minerals  and  How  to  Study  Them i2mo,  1  50 

Catalogue  of  American  Localities  of  Minerals Large  8vo,  i  00 

Manual  of  Mineralogy  and  Petrography i2mo  2  00 

Douglas's  Untechnical  Addresses  on  Technical  Subjects i2mo,  i  00 

Eakle's  Mineral  Tables 8vo,  i  25 

Egleston's  Catalogue  of  Minerals  and  Synonyms 8vo,  2  50 

Hussak's  The  Determination  of  Rock-forming  Minerals.    (Smith.). Small  Svo,  2  00 

Merrill's  Non-metallic  Minerals;   Their  Occurrence  and  Uses Svo,  4  00 

*  Penfield's  Notes  on  Determinative  Mineralogy  and  Record  of  Mineral  Tests. 

Svo   paper,  o  50 
Rosenbusch's   Microscopical   Physiography   ot   the    Rock-making  Minerals 

(lddings.1 Svo.  5  00 

*  Tilhnans  Text-book  of  Important  Mmerals  and  Rocks Svo.  2  00 

Williams's  Manual  of  Lithology Svo,  3  00 

MINING. 

Beard's  Ventilation  of  Mines i2mo.  2  50 

Boyd's  Resources  of  Southwest  Virginia Svo,  3  00 

Map  of  Southwest  Virginia Pocket  book  form.  2  00 

Douglas's  Untechnical  Addresses  on  Technical  Subjects i2mo.  i  00 

*  Drinker's  Tunneling,  Explosive  Compounds,  and  Rock  Drills.  .4to.hf  mor  25  00 

Eissler's  Modern  High  Explosives Svo  4  00 

Fowler's  Sewage  Works  Analyses i2mo  2  00 

Goodyear's  Coal-mines  of  the  Western  Coast  of  the  United  States i2mo.  2  50 

Ihlseng's  Manual  of  Mining . Svo.  5  00 

**  Iles's  Lead-smelting.     (Postage  gc.  additional.) i2mo.  2  50 

Kunhardt's  Practice  of  Ore  Dressing  in  Europe.    .    Svo,  1   50 

O'Driscoll's  Notes  on  the  Treatment  of  Gold  Ores Svo.  i  00 

*  Walke's  Lectures  on  Explosives Svo,    4  00 

Wilson's  Cyanide  Processes i2mo.,     i  50 

Chiorination  Process i2mo,    1  50 

15 


Wilson's  Hydraulic  and  Placer  Mining i2mo,  2  00 

Treatise  on  Practical  and  Theoretical  Mine  Ventilation i2mo,  i  25 

SANITARY  SCIENCE. 

Bashore's  Sanitation  of  a  Country  House i2mo,  i  00 

FolweU's  Sewerage.     (Designing,  Construction,  and  Maintenance.) 8vo,  3  00 

Water-supply  Engineering 8vo,  4  00 

Fuertes's  Water  and  Public  Health i2mo,  1  50 

Water-filtration  Works i2mo,  2  50 

Gerhard's  Guide  to  Sanitary  House-inspection i6mo,  1  00 

Goodrich's  Economic  Disposal  of  Town's  Refuse DemySvo,  3  50 

Hazen's  Filtration  of  Public  Water-supplies 8vo,  3  00 

Leach's  The  Inspection  and  Analysis  of  Food  with  Special  Reference  to  State 

Control 8vo,  7  50 

Mawu's  Water-supply.  (Considered  principallyfrom  a  Sanitary  Standpoint)  8vo,  4  00 

Examination  of  Water.     (Chemical  and  Bacteriological.) i2mo,  i  25 

Merriman's  Elements  of  Sanitary  Engineering 8vo,  2  00 

Ogden's  Sewer  Design i2mo,  2  00 

Prescott  and  Winslow's  Elements  of  Water  Bacteriology,  with  Special  Refer- 
ence to  Sanitary  Water  Analysis i2mo,  i  25 

*  Price's  Handbook  on  Sanitation i2mo,  i  50 

Richards's  Cost  of  Food.     A  Study  in  Dietaries i2mo,  i  00 

Cost  of  Living  as  Modified  by  Sanitaiy  Science i2mo,  i  o'j 

Richards  and  Woodman's  Air,  Water,  and  Food  from  a  Sanitary  Stand- 
point  8vo,  2  00 

*  Richards  and  Williams's  The  Dietary  Computer 8vo,  i  50 

Rideal's  Sewage  and  Bacterial  Purification  of  Sewage 8vo,  3  50 

Turneaure  and  Russell's  Public  Water-supplies 8vo,  5  00 

Von  Behring's  Suppression  of  Tuberculosis.     (Bolduan.) i2mo,  i  00 

Whipple's  Microscopy  of  Drinking-water 8vo,  3  50 

WoodhuU's  Notes  on  Military  Hygiene 16100,  i  50 

MISCELLANEOUS. 

De  Fursac's  Manual  of  Psychiatry.  fRosanoff  and  Collins.).  . .  .Large  i2mo,  a  50 
Emmons's  Geological  Guide-book  of  the  Rocky  Mountain  Excursion  of  the 

International  Congress  of  Geologists Large  8vo,  i  50 

Ferrel's  Popular  Treatise  on  the  Winds 8vo.  4  00 

Haines's  American  Railway  Management i2mo,  2  50 

Mott's  Composition,  Digestibility,  and  Nutritive  Value  of  Food.  Mounted  chart,  i  25 

Fallacy  of  the  Present  Theory  of  Sound i6mo,  i  00 

Ricketts's  History  of  Rensselaer  Polytechnic  Institute,  1824-1894.  .Small  8vo,  3  00 

Rostoski's  Serum  Diagnosis.     (Bolduan.) i2mo,  i  00 

Rotherham's  Emphasized  New  Testament Large  8vo,  2  00 

Steel's  Treatise  on  the  Diseases  of  the  Dog 8vo,  3  30 

Totten's  Important  Question  in  Metrology 8vo,  2  50 

The  World's  Columbian  Exposition  of  1893 4to,  i  00 

Von  Behring's  Suppression  of  Tuberculosis.     (Bolduan.) i2mo,  1  00 

Winslow's  Elements  of  Applied  Microscopy i2mo,  1  50 

Worcester  and  Atkinson.     Small  Hospitals,  Establishment  and  Maintenance; 

Suggestions  for  Hospital  Architecture :  Plans  for  Small  Hospital .  i2mo,  i  25 

HEBREW  AND   CHALDEE  TEXT-BOOKS. 

Green's  Elementary  Hebrew  Grammar i2mo,  i  25 

Hebrew  Chrestomathy 8vo,  a  00 

Gesenius's  Hebrew  and  Chaldee  Lexicon  tr    the  Old  Testament  Scriptures. 

(Tregelles.) Small  4to,  half  morocco,  5  00 

Letter4s's  Hebrew  Bible 8vo,  2  25 

16 


UNIVERSITY  OF  CALIFORNIA  LIBRARY 

Los  Angeles  |\,\jA\_.r  >  ' 

This  book  is  DUE  on  the  last  date  stamped  below.  S^^^<^■ 


<0   - 

OCT 


APR  -1  21965 
^^B  2  81956 


HH^U  ^955 


APR  ^  ^  ^^^,,c.^ 


OCT  X^ 


OCT  1  3  REC0 

JAN  2  2  RECB 


i/'ay 


if9egf 


^^^6^ 


OCT  2f^89 
OCT  2  0  ^^^'^ 


N 


^4  198' 


,       RLC'DAUPL. 

MAR  06  199] 
tEB  2  6  1991 

EMS  LIBRARY 


Form  L9-lC0m-9,'52(A3105)444 


Uhrary 

Uo4 


A     000  351  042     7 


r^'il'-:!^' 


1' 


iT"fC 


''■-.  V.    •^' 


'-mm::-, 

,v;;;^v<A|Jf^;:';,  , 


|;N^''!{^vN^. 


